Jika kamu ingin mempelajarinya bentuk akar istimewa secara lebih mendalam, coba simak penjelasan yang ada di sini. Setelah menerima materi, kamu bisa langsung mempraktikkannya dengan mengerjakan latihan soal yang telah kami sediakan.
Di sini, kamu akan belajar tentang Bentuk Akar Istimewa melalui video yang dibawakan oleh Bapak Anton Wardaya. Kamu akan diajak untuk memahami materi hingga metode menyelesaikan soal.
Selain itu, kamu juga akan mendapatkan latihan soal interaktif dalam 3 tingkat kesulitan (mudah, sedang, sukar). Dengan begitu, kamu bisa langsung mempraktikkan materi yang telah dijelaskan.
Sekarang, kamu bisa mulai belajar dengan 3 video dan 3 set latihan soal yang ada di halaman ini. Apabila materi ini berguna, bagikan ke teman atau rekan kamu supaya mereka juga mendapatkan manfaatnya.
Kamu dapat download modul & contoh soal serta kumpulan latihan soal lengkap dalam bentuk pdf pada list dibawah ini:
Bentuk Akar Istimewa (1)
Bentuk Akar Istimewa (2)
Contoh Soal Bentuk Akar Istimewa
Latihan Soal Bentuk Akar Istimewa (Mudah)
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Pertanyaan ke 1 dari 5
1. Pertanyaan
Bentuk sederhana dari $\sqrt{\left(a^{3}+a^{2}\right)}$ adalah…
Betul$\begin{aligned}\sqrt{\left(a^{3}+a^{2}\right)} & =\sqrt{a^{2}\left(a+1\right)}\\
& =a\sqrt{\left(a+1\right)}
\end{aligned}
$Salah$\begin{aligned}\sqrt{\left(a^{3}+a^{2}\right)} & =\sqrt{a^{2}\left(a+1\right)}\\
& =a\sqrt{\left(a+1\right)}
\end{aligned}
$ -
Pertanyaan ke 2 dari 5
2. Pertanyaan
Bentuk pangkat positif dari $\left(\frac{x^{-2}-y^{-2}}{y-x}\right)$ adalah…
Betul$\begin{aligned}\left(\frac{x^{-2}-y^{-2}}{y-x}\right) & =\frac{\frac{1}{x^{2}}-\frac{1}{y^{2}}}{y-x}\\
& =\frac{\frac{y^{2}-x^{2}}{x^{2}y^{2}}}{y-x}\\
& =\frac{y^{2}-x^{2}}{x^{2}y^{2}(y-x)}\\
& =\frac{(y+x)(y-x)}{x^{2}y^{2}(y-x)}\\
& =\frac{x+y}{x^{2}y^{2}}
\end{aligned}
$Salah$\begin{aligned}\left(\frac{x^{-2}-y^{-2}}{y-x}\right) & =\frac{\frac{1}{x^{2}}-\frac{1}{y^{2}}}{y-x}\\
& =\frac{\frac{y^{2}-x^{2}}{x^{2}y^{2}}}{y-x}\\
& =\frac{y^{2}-x^{2}}{x^{2}y^{2}(y-x)}\\
& =\frac{(y+x)(y-x)}{x^{2}y^{2}(y-x)}\\
& =\frac{x+y}{x^{2}y^{2}}
\end{aligned}
$ -
Pertanyaan ke 3 dari 5
3. Pertanyaan
Bentuk sederhana dari $\frac{x^{-1}y-xy^{-1}}{x^{-1}-y^{-1}}$ adalah….
Betul$\begin{aligned}\frac{x^{-1}y-xy^{-1}}{x^{-1}-y^{-1}}=\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{x}-\frac{1}{y}} & =\frac{\frac{y^{2}-x^{2}}{xy}}{\frac{y-x}{xy}}\\
& =\frac{(y+x)(y-x)}{y-x}\\
& =x+y
\end{aligned}
$Salah$\begin{aligned}\frac{x^{-1}y-xy^{-1}}{x^{-1}-y^{-1}}=\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{x}-\frac{1}{y}} & =\frac{\frac{y^{2}-x^{2}}{xy}}{\frac{y-x}{xy}}\\
& =\frac{(y+x)(y-x)}{y-x}\\
& =x+y
\end{aligned}
$ -
Pertanyaan ke 4 dari 5
4. Pertanyaan
Bentuk pangkat positif dari $\frac{x^{-4}-y^{-4}}{x^{-2}+y^{-2}}$ adalah….
Betul$\begin{aligned}\frac{x^{-4}-y^{-4}}{x^{-2}+y^{-2}} & =\frac{\left(x^{-2}\right)^{2}-\left(y^{-2}\right)^{2}}{x^{-2}+y^{-2}}\\
& =\frac{(x^{-2}+y^{-2})(x^{-2}-y^{-2})}{x^{-2}+y^{-2}}\\
& =x^{-2}-y^{-2}\\
& =\frac{1}{x^{2}}-\frac{1}{y^{2}}\\
& =\frac{y^{2}-x^{2}}{x^{2}y^{2}}
\end{aligned}
$Salah$\begin{aligned}\frac{x^{-4}-y^{-4}}{x^{-2}+y^{-2}} & =\frac{\left(x^{-2}\right)^{2}-\left(y^{-2}\right)^{2}}{x^{-2}+y^{-2}}\\
& =\frac{(x^{-2}+y^{-2})(x^{-2}-y^{-2})}{x^{-2}+y^{-2}}\\
& =x^{-2}-y^{-2}\\
& =\frac{1}{x^{2}}-\frac{1}{y^{2}}\\
& =\frac{y^{2}-x^{2}}{x^{2}y^{2}}
\end{aligned}
$ -
Pertanyaan ke 5 dari 5
5. Pertanyaan
Jika $xy=7$ , maka $\frac{2^{(x+y)^{2}}}{2^{(x-y)^{2}}}$ adalah….
BetulDiketahui $xy=7$
$\begin{aligned}\frac{2^{(x+y)^{2}}}{2^{(x-y)^{2}}} & =2^{(x+y)^{2}-\left(x-y\right)^{2}}\\
& =2^{x^{2+2xy+y^{2}-x^{2}+2xy-y^{2}}}\\
& =2^{4xy}\\
& =2^{4(7)}\\
& =2^{28}
\end{aligned}
$SalahDiketahui $xy=7$
$\begin{aligned}\frac{2^{(x+y)^{2}}}{2^{(x-y)^{2}}} & =2^{(x+y)^{2}-\left(x-y\right)^{2}}\\
& =2^{x^{2+2xy+y^{2}-x^{2}+2xy-y^{2}}}\\
& =2^{4xy}\\
& =2^{4(7)}\\
& =2^{28}
\end{aligned}
$
Latihan Soal Bentuk Akar Istimewa (Sedang)
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1. Pertanyaan
Bentuk Sederhana dari $\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)^{2}-\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)$$^{2}$ adalah…
Betul$\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)^{2}-\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)^{2}$
$=\left[\sqrt{2}+\left(\sqrt{3}-\sqrt{6}\right)\right]^{2}-\left[\sqrt{2}-\left(\sqrt{3}-\sqrt{6}\right)\right]^{2}$
$=2+2\sqrt{2}\left(\sqrt{3}-\sqrt{6}\right)+\left(\sqrt{3}-\sqrt{6}\right)^{2}-2$$+2\sqrt{2}\left(\sqrt{3}-\sqrt{6}\right)-\left(\sqrt{3}-\sqrt{6}\right)^{2}$
$=4\sqrt{2}\left(\sqrt{3}-\sqrt{6}\right)$
$=4\sqrt{2}\sqrt{3}(1-\sqrt{2})$
$=4\sqrt{6}(1-\sqrt{2})$
Salah$\left(\sqrt{2}+\sqrt{3}-\sqrt{6}\right)^{2}-\left(\sqrt{2}-\sqrt{3}+\sqrt{6}\right)^{2}$
$=\left[\sqrt{2}+\left(\sqrt{3}-\sqrt{6}\right)\right]^{2}-\left[\sqrt{2}-\left(\sqrt{3}-\sqrt{6}\right)\right]^{2}$
$=2+2\sqrt{2}\left(\sqrt{3}-\sqrt{6}\right)+\left(\sqrt{3}-\sqrt{6}\right)^{2}-2$$+2\sqrt{2}\left(\sqrt{3}-\sqrt{6}\right)-\left(\sqrt{3}-\sqrt{6}\right)^{2}$
$=4\sqrt{2}\left(\sqrt{3}-\sqrt{6}\right)$
$=4\sqrt{2}\sqrt{3}(1-\sqrt{2})$
$=4\sqrt{6}(1-\sqrt{2})$
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Pertanyaan ke 2 dari 5
2. Pertanyaan
Akar-Akar persamaan $9^{2x^{2}-6x+1}=27^{2x-4}$adalah $\alpha$ dan $\beta$. Nilai $\alpha$+$\beta$ adalah…
Betul$9^{2x^{2}-6x+1}=27^{2x-4}$
$3^{2(2x^{2}-6x+1)}=3^{3(2x-4)}$$\rightarrow2(2x^{2}-6x+1)=3(2x-4)$
$4x^{2}-12x+2=6x-12$
$4x^{2}-18x+14=0$
Jadi $\alpha$+$\beta=-\frac{b}{a}=-\frac{(-18)}{4}=\frac{9}{2}=4\frac{1}{2}$
Salah$9^{2x^{2}-6x+1}=27^{2x-4}$
$3^{2(2x^{2}-6x+1)}=3^{3(2x-4)}$$\rightarrow2(2x^{2}-6x+1)=3(2x-4)$
$4x^{2}-12x+2=6x-12$
$4x^{2}-18x+14=0$
Jadi $\alpha$+$\beta=-\frac{b}{a}=-\frac{(-18)}{4}=\frac{9}{2}=4\frac{1}{2}$
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Pertanyaan ke 3 dari 5
3. Pertanyaan
Bentuk sederhana dari $\frac{ab^{-1}-a^{-1}b}{b^{-2}-a^{-2}}$ adalah…
Betul$\begin{aligned}\frac{ab^{-1}-a^{-1}b}{b^{-2}-a^{-2}} & =\frac{\frac{a}{b}-\frac{b}{a}}{\frac{1}{b^{2}}-\frac{1}{a^{2}}}\\
& =\frac{\frac{a^{2}-b^{2}}{ab}}{\frac{a^{2}-b^{2}}{\left(ab\right)^{2}}}\\
& =ab
\end{aligned}
$Salah$\begin{aligned}\frac{ab^{-1}-a^{-1}b}{b^{-2}-a^{-2}} & =\frac{\frac{a}{b}-\frac{b}{a}}{\frac{1}{b^{2}}-\frac{1}{a^{2}}}\\
& =\frac{\frac{a^{2}-b^{2}}{ab}}{\frac{a^{2}-b^{2}}{\left(ab\right)^{2}}}\\
& =ab
\end{aligned}
$ -
Pertanyaan ke 4 dari 5
4. Pertanyaan
Jika $a=\frac{1}{2}\left(e^{x}-e^{-x}\right)$dan $b=\frac{1}{2}\left(e^{x}+e^{-x}\right)$, maka $\left(a^{2}-b^{2}\right)^{2}$ adalah…
Betul$a=\frac{1}{2}\left(e^{x}-e^{-x}\right)$ dan $b=\frac{1}{2}\left(e^{x}+e^{-x}\right)$
$a^{2}=\frac{1}{4}\left(e^{2x}-2+e^{-2x}\right)$ dan $b^{2}=\frac{1}{4}\left(e^{2x}+2+e^{-2x}\right)$
$\begin{aligned}a^{2}-b^{2} & =\frac{1}{4}\left(e^{2x}-2+e^{-2x}-e^{2x}-2-e^{-2x}\right)\\
& =\frac{1}{4}(-4)\\
& =-1
\end{aligned}
$Jadi $\begin{aligned}\left(a^{2}-b^{2}\right)^{2} & =(-1)^{2}\\
& =1
\end{aligned}
$Salah$a=\frac{1}{2}\left(e^{x}-e^{-x}\right)$ dan $b=\frac{1}{2}\left(e^{x}+e^{-x}\right)$
$a^{2}=\frac{1}{4}\left(e^{2x}-2+e^{-2x}\right)$ dan $b^{2}=\frac{1}{4}\left(e^{2x}+2+e^{-2x}\right)$
$\begin{aligned}a^{2}-b^{2} & =\frac{1}{4}\left(e^{2x}-2+e^{-2x}-e^{2x}-2-e^{-2x}\right)\\
& =\frac{1}{4}(-4)\\
& =-1
\end{aligned}
$Jadi $\begin{aligned}\left(a^{2}-b^{2}\right)^{2} & =(-1)^{2}\\
& =1
\end{aligned}
$ -
Pertanyaan ke 5 dari 5
5. Pertanyaan
Jika $a=\frac{3-\sqrt{2}}{3+\sqrt{2}}$ dan $b=\frac{3+\sqrt{2}}{3-\sqrt{2}}$, nilai dari $a^{2}-b^{2}$adalah…
Betul$a^{2}-b^{2}=\left(a+b\right)(a-b)$
$\begin{aligned}a & =\frac{3-\sqrt{2}}{3+\sqrt{2}}\times\frac{3-\sqrt{2}}{3-\sqrt{2}}\\
& =\frac{\left(3-\sqrt{2}\right)^{2}}{9-2}\\
& =\frac{\left(3-\sqrt{2}\right)^{2}}{7}
\end{aligned}
$$\begin{aligned}b & =\frac{3+\sqrt{2}}{3-\sqrt{2}}\times\frac{3+\sqrt{2}}{3+\sqrt{2}}\\
& =\frac{\left(3+\sqrt{2}\right)^{2}}{9-2}\\
& =\frac{\left(3+\sqrt{2}\right)^{2}}{7}
\end{aligned}
$$a^{2}-b^{2}=\left[\frac{\left(3-\sqrt{2}\right)^{2}}{7}+\frac{\left(3+\sqrt{2}\right)^{2}}{7}\right]\left[\frac{\left(3-\sqrt{2}\right)^{2}}{7}-\frac{\left(3+\sqrt{2}\right)^{2}}{7}\right]$
$=\frac{1}{7}\left[\left(3-\sqrt{2}\right)^{2}+\left(3+\sqrt{2}\right)^{2}\right]$$\frac{1}{7}\left[\left(3-\sqrt{2}\right)^{2}-\left(3+\sqrt{2}\right)^{2}\right]$
$=\frac{1}{49}\left[9-6\sqrt{2}+2+9+6\sqrt{2}+2\right]$$\left[9-6\sqrt{2}+2-9-6\sqrt{2}-2\right]$
$=\frac{1}{49}\left(22\right)(-12\sqrt{2})$
$=-\frac{264}{49}\sqrt{2}$
Salah$a^{2}-b^{2}=\left(a+b\right)(a-b)$
$\begin{aligned}a & =\frac{3-\sqrt{2}}{3+\sqrt{2}}\times\frac{3-\sqrt{2}}{3-\sqrt{2}}\\
& =\frac{\left(3-\sqrt{2}\right)^{2}}{9-2}\\
& =\frac{\left(3-\sqrt{2}\right)^{2}}{7}
\end{aligned}
$$\begin{aligned}b & =\frac{3+\sqrt{2}}{3-\sqrt{2}}\times\frac{3+\sqrt{2}}{3+\sqrt{2}}\\
& =\frac{\left(3+\sqrt{2}\right)^{2}}{9-2}\\
& =\frac{\left(3+\sqrt{2}\right)^{2}}{7}
\end{aligned}
$$a^{2}-b^{2}=\left[\frac{\left(3-\sqrt{2}\right)^{2}}{7}+\frac{\left(3+\sqrt{2}\right)^{2}}{7}\right]\left[\frac{\left(3-\sqrt{2}\right)^{2}}{7}-\frac{\left(3+\sqrt{2}\right)^{2}}{7}\right]$
$=\frac{1}{7}\left[\left(3-\sqrt{2}\right)^{2}+\left(3+\sqrt{2}\right)^{2}\right]$$\frac{1}{7}\left[\left(3-\sqrt{2}\right)^{2}-\left(3+\sqrt{2}\right)^{2}\right]$
$=\frac{1}{49}\left[9-6\sqrt{2}+2+9+6\sqrt{2}+2\right]$$\left[9-6\sqrt{2}+2-9-6\sqrt{2}-2\right]$
$=\frac{1}{49}\left(22\right)(-12\sqrt{2})$
$=-\frac{264}{49}\sqrt{2}$
Latihan Soal Bentuk Akar Istimewa (Sukar)
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Pertanyaan ke 1 dari 5
1. Pertanyaan
Bentuk sederhana dari $\frac{1}{\left(\sqrt{7}-\sqrt{5}\right)^{2}}-\frac{1}{\left(\sqrt{7}+\sqrt{5}\right)^{2}}$ adalah…
Betul$\frac{1}{\left(\sqrt{7}-\sqrt{5}\right)^{2}}-\frac{1}{\left(\sqrt{7}+\sqrt{5}\right)^{2}}$
$=\frac{\left(\sqrt{7}+\sqrt{5}\right)^{2}-\left(\sqrt{7}-\sqrt{5}\right)^{2}}{\left(\sqrt{7}-\sqrt{5}\right)^{2}\left(\sqrt{7}+\sqrt{5}\right)^{2}}$
$=\frac{\left(\sqrt{7}+\sqrt{5}+\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}\right)}{\left[\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\right]^{2}}$
$=\frac{2\sqrt{7}\,2\sqrt{5}}{4}=\sqrt{35}$
Salah$\frac{1}{\left(\sqrt{7}-\sqrt{5}\right)^{2}}-\frac{1}{\left(\sqrt{7}+\sqrt{5}\right)^{2}}$
$=\frac{\left(\sqrt{7}+\sqrt{5}\right)^{2}-\left(\sqrt{7}-\sqrt{5}\right)^{2}}{\left(\sqrt{7}-\sqrt{5}\right)^{2}\left(\sqrt{7}+\sqrt{5}\right)^{2}}$
$=\frac{\left(\sqrt{7}+\sqrt{5}+\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}\right)}{\left[\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\right]^{2}}$
$=\frac{2\sqrt{7}\,2\sqrt{5}}{4}=\sqrt{35}$
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Pertanyaan ke 2 dari 5
2. Pertanyaan
Bentuk sederhana dalam pangkat positif dari :
$\frac{\left[x\left(x^{2}-y^{2}\right)\right]^{3}.\left[y\left(x+y\right)\right]^{4}}{\left(x^{3}+y^{3}\right)^{4}.\left(x-y\right)^{3}}\cdot\frac{\left[x^{2}-xy+y^{2}\right]^{4}}{\left(xy\right)^{3}}$ adalah…
BetulIngat bentuk $\left(a+b\right)^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right)$
$\frac{x^{3}\left(x+y\right)^{3}\left(x-y\right)^{3}y^{4}\left(x+y\right)^{4}}{\left(x^{3}+y^{3}\right)^{4}\left(x-y\right)^{3}}\cdot\frac{\left(x^{2}-xy+y^{2}\right)^{4}}{x^{3}y^{3}}$
$=\frac{x^{3}\left(x+y\right)^{3}\left(x-y\right)^{3}y^{4}\left(x+y\right)^{4}}{\left[\left(x+y\right)\left(x^{2}-xy+y^{2}\right)\right]^{4}\left(x-y\right)^{3}}\cdot\frac{\left(x^{2}-xy+y^{2}\right)^{4}}{x^{3}y^{3}}$
$=\frac{x^{3}\left(x+y\right)^{3}\left(x-y\right)^{3}y^{4}.\left(x+y\right)^{4}}{\left(x+y\right)^{4}\left(x^{2}-xy+y^{2}\right)^{4}\left(x-y\right)^{3}}\cdot\frac{\left(x^{2}-xy+y^{2}\right)^{4}}{x^{3}y^{3}}$
$=y\left(x+y\right)^{3}$
SalahIngat bentuk $\left(a+b\right)^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right)$
$\frac{x^{3}\left(x+y\right)^{3}\left(x-y\right)^{3}y^{4}\left(x+y\right)^{4}}{\left(x^{3}+y^{3}\right)^{4}\left(x-y\right)^{3}}\cdot\frac{\left(x^{2}-xy+y^{2}\right)^{4}}{x^{3}y^{3}}$
$=\frac{x^{3}\left(x+y\right)^{3}\left(x-y\right)^{3}y^{4}\left(x+y\right)^{4}}{\left[\left(x+y\right)\left(x^{2}-xy+y^{2}\right)\right]^{4}\left(x-y\right)^{3}}\cdot\frac{\left(x^{2}-xy+y^{2}\right)^{4}}{x^{3}y^{3}}$
$=\frac{x^{3}\left(x+y\right)^{3}\left(x-y\right)^{3}y^{4}.\left(x+y\right)^{4}}{\left(x+y\right)^{4}\left(x^{2}-xy+y^{2}\right)^{4}\left(x-y\right)^{3}}\cdot\frac{\left(x^{2}-xy+y^{2}\right)^{4}}{x^{3}y^{3}}$
$=y\left(x+y\right)^{3}$
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Pertanyaan ke 3 dari 5
3. Pertanyaan
Bentuk sederhana dari $\frac{\left(\sqrt{8-3\sqrt{7}}\right)\left(3+\sqrt{7}\right)}{\sqrt{3+\sqrt{5}}}$ adalah…
Betul$\frac{\left(\sqrt{8-3\sqrt{7}}\right)\left(3+\sqrt{7}\right)}{\sqrt{3+\sqrt{5}}}=\frac{\left(\sqrt{8-2\sqrt{\frac{9}{4}.7}}\right)\left(3+\sqrt{7}\right)}{\sqrt{3+2\sqrt{\frac{5}{4}}}}$
$=\frac{\sqrt{\frac{1}{2}}\left(\sqrt{16-2\sqrt{63}}\right)\left(3+\sqrt{7}\right)}{\sqrt{\frac{1}{2}}\left(\sqrt{6+2\sqrt{5}}\right)}$
$=\frac{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\sqrt{5}+1}$
$=\frac{9-7}{\sqrt{5}+1}$
$=\frac{2}{\sqrt{5}+1}$
$=\frac{2}{\sqrt{5}+1}\times\frac{\sqrt{5}-1}{\sqrt{5}-1}$
$=\frac{2\left(\sqrt{5}-1\right)}{4}$
$=\frac{1}{2}\left(\sqrt{5}-1\right)$
Salah$\frac{\left(\sqrt{8-3\sqrt{7}}\right)\left(3+\sqrt{7}\right)}{\sqrt{3+\sqrt{5}}}=\frac{\left(\sqrt{8-2\sqrt{\frac{9}{4}.7}}\right)\left(3+\sqrt{7}\right)}{\sqrt{3+2\sqrt{\frac{5}{4}}}}$
$=\frac{\sqrt{\frac{1}{2}}\left(\sqrt{16-2\sqrt{63}}\right)\left(3+\sqrt{7}\right)}{\sqrt{\frac{1}{2}}\left(\sqrt{6+2\sqrt{5}}\right)}$
$=\frac{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\sqrt{5}+1}$
$=\frac{9-7}{\sqrt{5}+1}$
$=\frac{2}{\sqrt{5}+1}$
$=\frac{2}{\sqrt{5}+1}\times\frac{\sqrt{5}-1}{\sqrt{5}-1}$
$=\frac{2\left(\sqrt{5}-1\right)}{4}$
$=\frac{1}{2}\left(\sqrt{5}-1\right)$
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Pertanyaan ke 4 dari 5
4. Pertanyaan
Bentuk Sederhana dari $\sqrt{\frac{2}{5-2\sqrt{6}}}-\sqrt{\frac{2}{5+2\sqrt{6}}}$ adalah…
Betul$\sqrt{\frac{2}{5-2\sqrt{6}}}-\sqrt{\frac{2}{5+2\sqrt{6}}}=\frac{\sqrt{2}}{\sqrt{5-2\sqrt{6}}}-\frac{\sqrt{2}}{\sqrt{5+2\sqrt{6}}}$
$=\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)-\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{3-2}$
$=\frac{\sqrt{6}+2-\sqrt{6}+2}{1}$
$=4$
Salah$\sqrt{\frac{2}{5-2\sqrt{6}}}-\sqrt{\frac{2}{5+2\sqrt{6}}}=\frac{\sqrt{2}}{\sqrt{5-2\sqrt{6}}}-\frac{\sqrt{2}}{\sqrt{5+2\sqrt{6}}}$
$=\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)-\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)}{3-2}$
$=\frac{\sqrt{6}+2-\sqrt{6}+2}{1}$
$=4$
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Pertanyaan ke 5 dari 5
5. Pertanyaan
Bentuk sederhana dari $\sqrt{2-\sqrt{-2+2\sqrt{5}}}-\sqrt{2+\sqrt{-2+2\sqrt{5}}}$ adalah…
BetulMisal $p=\sqrt{2-\sqrt{-2+2\sqrt{5}}}-\sqrt{2+\sqrt{-2+2\sqrt{5}}}$, jika dikudratkan maka :
$p^{2}=\left(\sqrt{2-\sqrt{-2+2\sqrt{5}}}-\sqrt{2+\sqrt{-2+2\sqrt{5}}}\right)^{2}$
$p^{2}=2-\sqrt{-2+2\sqrt{5}}-2\sqrt{\left(2\right)^{2}-\left(\sqrt{-2+2\sqrt{5}}\right)^{2}}$$+2+\sqrt{-2+2\sqrt{5}}$
$p^{2}=4-2\sqrt{4-\left(-2+2\sqrt{5}\right)}$
$p^{2}=4-2\sqrt{6-2\sqrt{5}}$
$p^{2}=4-2\left(\sqrt{5}-1\right)=6-2\sqrt{5}$
$p=\pm\sqrt{6-2\sqrt{5}}$
$p=\pm\left(\sqrt{5}-1\right)$
Jadi $p=\sqrt{5}-1$ dan $p=-\sqrt{5}+1$
SalahMisal $p=\sqrt{2-\sqrt{-2+2\sqrt{5}}}-\sqrt{2+\sqrt{-2+2\sqrt{5}}}$, jika dikudratkan maka :
$p^{2}=\left(\sqrt{2-\sqrt{-2+2\sqrt{5}}}-\sqrt{2+\sqrt{-2+2\sqrt{5}}}\right)^{2}$
$p^{2}=2-\sqrt{-2+2\sqrt{5}}-2\sqrt{\left(2\right)^{2}-\left(\sqrt{-2+2\sqrt{5}}\right)^{2}}$$+2+\sqrt{-2+2\sqrt{5}}$
$p^{2}=4-2\sqrt{4-\left(-2+2\sqrt{5}\right)}$
$p^{2}=4-2\sqrt{6-2\sqrt{5}}$
$p^{2}=4-2\left(\sqrt{5}-1\right)=6-2\sqrt{5}$
$p=\pm\sqrt{6-2\sqrt{5}}$
$p=\pm\left(\sqrt{5}-1\right)$
Jadi $p=\sqrt{5}-1$ dan $p=-\sqrt{5}+1$