HOTS Zone : Bentuk Akar

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Tipe:

Standar SBMPTN HOTS

No.

Jika $\sqrt{9x^2-5x+950}+\sqrt{9x^2-5x-945}=379$, maka nilai $\sqrt{9x^2-5x+950}-\sqrt{9x^2-5x-945}$ adalah ....

OMVN 2018

ALTERNATIF PENYELESAIAN

Misal $a=\sqrt{9x^2-5x+950}$ dan $b=\sqrt{9x^2-5x-945}$
$$\begin{array}{rl}{a}^2-b^2& =(9x^2-5x+950)-(9x^2-5x-945)\\ (a+b)(a-b)& =9x^2-5x+950-9x^2+5x+945\\ 375(a-b)& =1895\\ a-b& =5\end{array}$$

Jadi, $\sqrt{9x^2-5x+950}-\sqrt{9x^2-5x-945}=5$.

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No.

Nilai dari $\sqrt{3200,5^2-3199,5^2}$ adalah

1. $\mathrm{82,123}$
2. $80$
3. $34$

4. $54$
5. $\mathrm{25,567}$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\sqrt{3200,5^2-3199,5^2}& =\sqrt{(3200,5+3199,5)(3200,5-3199,5)}\\ & =\sqrt{(6400)(1)}\\ & =80\end{array}$$

Jadi, $\sqrt{3200,5^2-3199,5^2}=80$.
JAWAB: B

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No.

${\displaystyle \frac{\sqrt{3^{2015}}}{\sqrt{3^{2015}}-\sqrt{3^{2013}}}}=$

1. ${\displaystyle \frac{3}{2}}$
2. ${\displaystyle \frac{4}{3}}$

3. ${\displaystyle \frac{5}{4}}$
4. ${\displaystyle \frac{6}{5}}$

Matematika Zone Id

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{\sqrt{3^{2015}}}{\sqrt{3^{2015}}-\sqrt{3^{2013}}}}& ={\displaystyle \frac{\sqrt{3^2\cdot 3^{2013}}}{\sqrt{3^2\cdot 3^{2013}}-\sqrt{3^{2013}}}}\\ & ={\displaystyle \frac{3\sqrt{3^{2013}}}{3\sqrt{3^{2013}}-\sqrt{3^{2013}}}}\\ & ={\displaystyle \frac{3\sqrt{3^{2013}}}{2\sqrt{3^{2013}}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{2}}}}}}\end{array}$$

Jadi, ${\displaystyle \frac{\sqrt{3^{2015}}}{\sqrt{3^{2015}}-\sqrt{3^{2013}}}}={\displaystyle \frac{3}{2}}$.
JAWAB: A

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No.

Jika $a$ dan $b$ bilangan asli dan $\sqrt{18+\sqrt{308}}=\sqrt{a}+\sqrt{b}$, maka nilai $a\times b$ adalah ....

OSN Guru Matematika SMA 2016

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\sqrt{18+\sqrt{308}}& =\sqrt{18+\sqrt{4\cdot 77}}\\ & =\sqrt{18+2\sqrt{77}}\\ & =\sqrt{a+b+2\sqrt{a\times b}}\end{array}$$
$a\times b=77$

Jadi, $a\times b=77$.

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No.

Tentukan semua bilangan bulat $n$ yang memenuhi $\sqrt{{\displaystyle \frac{25}{2}}+\sqrt{{\displaystyle \frac{625}{4}}-n}}+\sqrt{{\displaystyle \frac{25}{2}}-\sqrt{{\displaystyle \frac{625}{4}}-n}}$ adalah bilangan bulat

Baltic Way 1993

ALTERNATIF PENYELESAIAN

Kita lihat bahwa $n\le {\displaystyle \frac{625}{4}}$

$\begin{array}{rl}x& =\sqrt{{\displaystyle \frac{25}{2}}+\sqrt{{\displaystyle \frac{625}{4}}-n}}+\sqrt{{\displaystyle \frac{25}{2}}-\sqrt{{\displaystyle \frac{625}{4}}-n}}\\ & =\sqrt{{(\sqrt{{\displaystyle \frac{25}{2}}+\sqrt{{\displaystyle \frac{625}{4}}-n}}+\sqrt{{\displaystyle \frac{25}{2}}-\sqrt{{\displaystyle \frac{625}{4}}-n}})}^2}\\ & =\sqrt{{\displaystyle \frac{25}{2}}+\sqrt{{\displaystyle \frac{625}{4}}-n}+2\sqrt{({\displaystyle \frac{25}{2}}+\sqrt{{\displaystyle \frac{625}{4}}-n})({\displaystyle \frac{25}{2}}-\sqrt{{\displaystyle \frac{625}{4}}-n})}+{\displaystyle \frac{25}{2}}-\sqrt{{\displaystyle \frac{625}{4}}-n}}\\ & =\sqrt{25+2\sqrt{{\displaystyle \frac{625}{4}}-{\displaystyle \frac{625}{4}}+n}}\\ & =\sqrt{25+2\sqrt{n}}\ge \sqrt{25}=5\end{array}$

$n={\left({\displaystyle \frac{x^2-25}{2}}\right)}^2$

Jika $x$ genap maka $x^2-25$ ganjil, sehingga $n$ bukan bilangan bulat. Jadi $x$ haruslah ganjil. Agar $n\le {\displaystyle \frac{625}{4}}$ maka maksimal $x^2$ adalah $50$. Sehingga kita dapat $x=5$ dan $x=7$.
Untuk $x=5$, $n=0$
Untuk $x=7$, $n=144$

Jadi, $n=0$ dan $n=144$.

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No.

Tentukan nilai dari $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})$

Diktat Pembinaan Olimpiade

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})& =(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{7}+\sqrt{5}-\sqrt{6})(\sqrt{7}-(\sqrt{5}-\sqrt{6}))\\ & =({(\sqrt{5}+\sqrt{6})}^2-7)(7-{(\sqrt{5}-\sqrt{6})}^2)\\ & =(5+2\sqrt{30}+6-7)(7-(5-2\sqrt{30}+6))\\ & =(2\sqrt{30}+4)(2\sqrt{30}-4)\\ & =4(30)-16\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 104}}}}\end{array}$$

Jadi, $(\sqrt{5}+\sqrt{6}+\sqrt{7})(\sqrt{5}+\sqrt{6}-\sqrt{7})(\sqrt{5}-\sqrt{6}+\sqrt{7})(-\sqrt{5}+\sqrt{6}+\sqrt{7})=104$.

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No.

Tentukan nilai $X$ yang memenuhi $X=(3-\sqrt{5})\left(\sqrt{3+\sqrt{5}}\right)+(3+\sqrt{5})\left(\sqrt{3-\sqrt{5}}\right)$

Diktat Pembinaan Olimpiade

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\left(\sqrt{3+\sqrt{5}}\right)\left(\sqrt{3-\sqrt{5}}\right)& =9-5\\ & =4\end{array}$$ $$$$\begin{array}{rl}X& =(3-\sqrt{5})\left(\sqrt{3+\sqrt{5}}\right)+(3+\sqrt{5})\left(\sqrt{3-\sqrt{5}}\right)\\ & =(3-\sqrt{5}){\displaystyle \frac{4}{\sqrt{3-\sqrt{5}}}}+(3+\sqrt{5}){\displaystyle \frac{4}{\sqrt{3+\sqrt{5}}}}\\ & =4\left(\sqrt{3-\sqrt{5}}\right)+4\left(\sqrt{3+\sqrt{5}}\right)\\ & =2\sqrt{2}\left(\sqrt{6-2\sqrt{5}}\right)+2\sqrt{2}\left(\sqrt{6+2\sqrt{5}}\right)\\ & =2\sqrt{2}(\sqrt{5}-1)+2\sqrt{2}(\sqrt{5}+1)\\ & =2\sqrt{10}-2\sqrt{2}+2\sqrt{10}+2\sqrt{2}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4\sqrt{10}}}}}\end{array}$$

Jadi, $X=4\sqrt{10}$.

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No.

Jika diketahui bahwa $\sqrt{14y^2-20y+48}+\sqrt{14y^2-20y-15}=9$, maka tentukan nilai dari $\sqrt{14y^2-20y+48}-\sqrt{14y^2-20y-15}=$

Diktat Pembinaan Olimpiade

ALTERNATIF PENYELESAIAN

Misal $a=\sqrt{14y^2-20y+48}$ dan $b=\sqrt{14y^2-20y-15}$
$$\begin{array}{rl}{a}^2-b^2& =(14y^2-20y+48)-(14y^2-20y-15)\\ (a+b)(a-b)& =14y^2-20y+48-14y^2+20y+15\\ 9(a-b)& =63\\ a-b& =\boxed{{\displaystyle \boxed{{\displaystyle 7}}}}\end{array}$$

Jadi, $\sqrt{14y^2-20y+48}-\sqrt{14y^2-20y-15}=7$.

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