HOTS Zone : Bentuk Akar
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Tipe:
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 ....ALTERNATIF PENYELESAIAN
Misal \(a=\sqrt{9x^2-5x+950}\) dan \(b=\sqrt{9x^2-5x-945}\)
\(\begin{aligned} a^2-b^2&=\left(9x^2-5x+950\right)-\left(9x^2-5x-945\right)\\ (a+b)(a-b)&=9x^2-5x+950-9x^2+5x+945\\ 375(a-b)&=1895\\ a-b&=5 \end{aligned}\)
\(\begin{aligned} a^2-b^2&=\left(9x^2-5x+950\right)-\left(9x^2-5x-945\right)\\ (a+b)(a-b)&=9x^2-5x+950-9x^2+5x+945\\ 375(a-b)&=1895\\ a-b&=5 \end{aligned}\)
Jadi, \(\sqrt{9x^2-5x+950}-\sqrt{9x^2-5x-945}=5\).
No.
Nilai dari \(\sqrt{3200{,}5^2-3199{,}5^2}\) adalah- 82,123
- 80
- 34
- 54
- 25,567
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\sqrt{3200{,}5^2-3199{,}5^2}&=\sqrt{(3200{,}5+3199{,}5)(3200{,}5-3199{,}5)}\\
&=\sqrt{(6400)(1)}\\
&=80
\end{aligned}\)
Jadi, \(\sqrt{3200{,}5^2-3199{,}5^2}=80\).
JAWAB: B
JAWAB: B
No.
- \(\dfrac32\)
- \(\dfrac43\)
- \(\dfrac54\)
- \(\dfrac65\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{\sqrt{3^{2015}}}{\sqrt{3^{2015}}-\sqrt{3^{2013}}}&=\dfrac{\sqrt{3^2\cdot3^{2013}}}{\sqrt{3^2\cdot3^{2013}}-\sqrt{3^{2013}}}\\[8pt]
&=\dfrac{3\sqrt{3^{2013}}}{3\sqrt{3^{2013}}-\sqrt{3^{2013}}}\\[8pt]
&=\dfrac{3\sqrt{3^{2013}}}{2\sqrt{3^{2013}}}\\
&=\boxed{\boxed{\dfrac32}}
\end{aligned}\)
Jadi, \(\dfrac{\sqrt{3^{2015}}}{\sqrt{3^{2015}}-\sqrt{3^{2013}}}=\dfrac32\).
JAWAB: A
JAWAB: A
No.
Jika a dan b bilangan asli dan \(\sqrt{18+\sqrt{308}}=\sqrt{a}+\sqrt{b}\), maka nilai a×b adalah ....ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\sqrt{18+\sqrt{308}}&=\sqrt{18+\sqrt{4\cdot77}}\\
&=\sqrt{18+2\sqrt{77}}\\
&=\sqrt{a+b+2\sqrt{a\times b}}
\end{aligned}\)
a×b = 77
a×b = 77
Jadi, a×b = 77.
No.
Tentukan semua bilangan bulat n yang memenuhi \(\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}}+\sqrt{\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\) adalah bilangan bulatALTERNATIF PENYELESAIAN
Kita lihat bahwa \(n\leq\dfrac{625}4\)
\(\begin{aligned} x&=\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}}+\sqrt{\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\\ &=\sqrt{\left(\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}}+\sqrt{\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\right)^2}\\ &=\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}+2\sqrt{\left(\dfrac{25}2+\sqrt{\dfrac{625}4-n}\right)\left(\dfrac{25}2-\sqrt{\dfrac{625}4-n}\right)}+\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\\ &=\sqrt{25+2\sqrt{\dfrac{625}4-\dfrac{625}4+n}}\\ &=\sqrt{25+2\sqrt{n}}\geq\sqrt{25}=5 \end{aligned}\)
\(n=\left(\dfrac{x^2-25}2\right)^2\)
Jika x genap maka x2 − 25 ganjil, sehingga n bukan bilangan bulat. Jadi x haruslah ganjil. Agar \(n\leq\dfrac{625}4\) maka maksimal x2 adalah 50. Sehingga kita dapat x = 5 dan x = 7.
Untuk x = 5, n = 0
Untuk x = 7, n = 144
\(\begin{aligned} x&=\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}}+\sqrt{\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\\ &=\sqrt{\left(\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}}+\sqrt{\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\right)^2}\\ &=\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}+2\sqrt{\left(\dfrac{25}2+\sqrt{\dfrac{625}4-n}\right)\left(\dfrac{25}2-\sqrt{\dfrac{625}4-n}\right)}+\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\\ &=\sqrt{25+2\sqrt{\dfrac{625}4-\dfrac{625}4+n}}\\ &=\sqrt{25+2\sqrt{n}}\geq\sqrt{25}=5 \end{aligned}\)
\(n=\left(\dfrac{x^2-25}2\right)^2\)
Jika x genap maka x2 − 25 ganjil, sehingga n bukan bilangan bulat. Jadi x haruslah ganjil. Agar \(n\leq\dfrac{625}4\) maka maksimal x2 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.
No.
Tentukan nilai dari\(\left(\sqrt5+\sqrt6+\sqrt7\right)\left(\sqrt5+\sqrt6-\sqrt7\right)\left(\sqrt5-\sqrt6+\sqrt7\right)\left(-\sqrt5+\sqrt6+\sqrt7\right)\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(\sqrt5+\sqrt6+\sqrt7\right)\left(\sqrt5+\sqrt6-\sqrt7\right)\left(\sqrt5-\sqrt6+\sqrt7\right)\left(-\sqrt5+\sqrt6+\sqrt7\right)&=\left(\sqrt5+\sqrt6+\sqrt7\right)\left(\sqrt5+\sqrt6-\sqrt7\right)\left(\sqrt7+\sqrt5-\sqrt6\right)\left(\sqrt7-(\sqrt5-\sqrt6)\right)\\
&=\left(\left(\sqrt5+\sqrt6\right)^2-7\right)\left(7-\left(\sqrt5-\sqrt6\right)^2\right)\\
&=\left(5+2\sqrt{30}+6-7\right)\left(7-\left(5-2\sqrt{30}+6\right)\right)\\
&=\left(2\sqrt{30}+4\right)\left(2\sqrt{30}-4\right)\\
&=4(30)-16\\
&=\boxed{\boxed{104}}
\end{aligned}\)
Jadi, \(\left(\sqrt5+\sqrt6+\sqrt7\right)\left(\sqrt5+\sqrt6-\sqrt7\right)\left(\sqrt5-\sqrt6+\sqrt7\right)\left(-\sqrt5+\sqrt6+\sqrt7\right)=104\).
No.
Tentukan nilai X yang memenuhi\(X=\left(3-\sqrt5\right)\left(\sqrt{3+\sqrt5}\right)+\left(3+\sqrt5\right)\left(\sqrt{3-\sqrt5}\right)\)
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(\sqrt{3+\sqrt5}\right)\left(\sqrt{3-\sqrt5}\right)&=9-5\\
&=4
\end{aligned}
\begin{aligned}
X&=\left(3-\sqrt5\right)\left(\sqrt{3+\sqrt5}\right)+\left(3+\sqrt5\right)\left(\sqrt{3-\sqrt5}\right)\\
&=\left(3-\sqrt5\right)\dfrac4{\sqrt{3-\sqrt5}}+\left(3+\sqrt5\right)\dfrac4{\sqrt{3+\sqrt5}}\\
&=4\left(\sqrt{3-\sqrt5}\right)+4\left(\sqrt{3+\sqrt5}\right)\\
&=2\sqrt2\left(\sqrt{6-2\sqrt5}\right)+2\sqrt2\left(\sqrt{6+2\sqrt5}\right)\\
&=2\sqrt2\left(\sqrt5-1\right)+2\sqrt2\left(\sqrt5+1\right)\\
&=2\sqrt{10}-2\sqrt2+2\sqrt{10}+2\sqrt2\\
&=\boxed{\boxed{4\sqrt{10}}}
\end{aligned}\)
Jadi, \(X=4\sqrt{10}\).
No.
Jika diketahui bahwa \({\sqrt{14y^2-20y+48}+\sqrt{14y^2-20y-15}=9}\), maka tentukan nilai dari \({\sqrt{14y^2-20y+48}-\sqrt{14y^2-20y-15}=}\)ALTERNATIF PENYELESAIAN
Misal \(a=\sqrt{14y^2-20y+48}\) dan \(b=\sqrt{14y^2-20y-15}\)
\(\begin{aligned} a^2-b^2&=\left(14y^2-20y+48\right)-\left(14y^2-20y-15\right)\\ (a+b)(a-b)&=14y^2-20y+48-14y^2+20y+15\\ 9(a-b)&=63\\ a-b&=\boxed{\boxed{7}} \end{aligned}\)
\(\begin{aligned} a^2-b^2&=\left(14y^2-20y+48\right)-\left(14y^2-20y-15\right)\\ (a+b)(a-b)&=14y^2-20y+48-14y^2+20y+15\\ 9(a-b)&=63\\ a-b&=\boxed{\boxed{7}} \end{aligned}\)
Jadi, \(\sqrt{14y^2-20y+48}-\sqrt{14y^2-20y-15}=7\).
No.
Nilai maksimum dari \({\sqrt{\left|x-y\right|}+\sqrt{\left|y-z\right|}+\sqrt{\left|z-x\right|}}\) dengan- \(\sqrt{6063}\)
- \(2\sqrt{2021}\)
- \(\sqrt{4042}+\sqrt{2021}\)
- \(2\sqrt{2021}+\sqrt{4042}\)
- \(2021\sqrt{2}+2021\)
ALTERNATIF PENYELESAIAN
Misalkan x ≥ y ≥ z dan \({P=\sqrt{\left|x-y\right|}+\sqrt{\left|y-z\right|}+\sqrt{\left|z-x\right|}}\). P bisa kita tulis menjadi,
\({P=\sqrt{x-y}+\sqrt{y-z}+\sqrt{x-z}}\)
Nilai P akan semakin besar jika nilai x semakin besar dan nilai z semakin kecil. Didapatx = 2021 dan z = 0 . Substitusikan ke P menjadi
\({P=\sqrt{2021-y}+\sqrt{y}+\sqrt{2021}}\)
Dengan menggunakan ketaksamaan Cauchy-Schwarz didapat,
\(\begin{aligned} \left(\sqrt{2021-y}\cdot1+\sqrt{y}\cdot1\right)^2&\leq \left(\left(\sqrt{2021-y}\right)^2+\left(\sqrt{y}\right)^2\right)\left(1^2+1^2\right)\\ \left(\sqrt{2021-y}+\sqrt{y}\right)^2&\leq \left(2021-y+y\right)\left(2\right)\\ \left(\sqrt{2021-y}+\sqrt{y}\right)^2&\leq 4042\\ \sqrt{2021-y}+\sqrt{y}&\leq \sqrt{4042}\\ \sqrt{2021-y}+\sqrt{y}+\sqrt{2021}&\leq \sqrt{4042}+\sqrt{2021} \end{aligned}\)
\({P=\sqrt{x-y}+\sqrt{y-z}+\sqrt{x-z}}\)
Nilai P akan semakin besar jika nilai x semakin besar dan nilai z semakin kecil. Didapat
\({P=\sqrt{2021-y}+\sqrt{y}+\sqrt{2021}}\)
Dengan menggunakan ketaksamaan Cauchy-Schwarz didapat,
\(\begin{aligned} \left(\sqrt{2021-y}\cdot1+\sqrt{y}\cdot1\right)^2&\leq \left(\left(\sqrt{2021-y}\right)^2+\left(\sqrt{y}\right)^2\right)\left(1^2+1^2\right)\\ \left(\sqrt{2021-y}+\sqrt{y}\right)^2&\leq \left(2021-y+y\right)\left(2\right)\\ \left(\sqrt{2021-y}+\sqrt{y}\right)^2&\leq 4042\\ \sqrt{2021-y}+\sqrt{y}&\leq \sqrt{4042}\\ \sqrt{2021-y}+\sqrt{y}+\sqrt{2021}&\leq \sqrt{4042}+\sqrt{2021} \end{aligned}\)
Jadi, nilai maksimum dari \({\sqrt{\left|x-y\right|}+\sqrt{\left|y-z\right|}+\sqrt{\left|z-x\right|}}\) dengan x,y,z ∈ [0, 2021] adalah \(\sqrt{4042}+\sqrt{2021}\).
JAWAB: C
JAWAB: C
No.
Diketahui\[2016\sqrt{(x+2016)(x-2016)}+2017\sqrt{(y+2017)(y-2017)}=\dfrac12\left(x^2+y^2\right)\]
maka nilai xy = ....
- 4066272
- 4068289
- 5750577,011
- 5756281,95
- 8132544
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
2016\sqrt{(x+2016)(x-2016)}+2017\sqrt{(y+2017)(y-2017)}&=\dfrac12\left(x^2+y^2\right)\\
2\cdot2016\sqrt{x^2-2106^2}+2\cdot2017\sqrt{y^2-2017^2}&=x^2-2016^2+2016^2+y^2-2017^2+2017^2
\end{aligned}\)
Misal \({\sqrt{x^2-2016^2}=p}\) dan \({\sqrt{y^2-2017^2}=q}\)
\(\begin{aligned} 2\cdot2016p+2\cdot2017q&=p^2+2016^2+q^2+2017^2\\ p^2-2\cdot2016p+2016^2+q^2-2\cdot2017q+2017^2&=0\\ (p-2016)^2+(q-2017)^2&=0 \end{aligned}\)
\(\begin{aligned} p-2016&=0\\ p&=2016\\ \sqrt{x^2-2016^2}&=2016\\ x^2-2016^2&=2016^2\\ x^2&=2\cdot2016^2\\ x&=2016\sqrt2 \end{aligned}\)
\(\begin{aligned} q-2017&=0\\ q&=2017\\ \sqrt{y^2-2017^2}&=2017\\ y&=2017\sqrt2 \end{aligned}\)
\(\begin{aligned} xy&=2016\sqrt2\cdot2017\sqrt2\\ &=\boxed{\boxed{8132544}} \end{aligned}\)
Misal \({\sqrt{x^2-2016^2}=p}\) dan \({\sqrt{y^2-2017^2}=q}\)
\(\begin{aligned} 2\cdot2016p+2\cdot2017q&=p^2+2016^2+q^2+2017^2\\ p^2-2\cdot2016p+2016^2+q^2-2\cdot2017q+2017^2&=0\\ (p-2016)^2+(q-2017)^2&=0 \end{aligned}\)
\(\begin{aligned} p-2016&=0\\ p&=2016\\ \sqrt{x^2-2016^2}&=2016\\ x^2-2016^2&=2016^2\\ x^2&=2\cdot2016^2\\ x&=2016\sqrt2 \end{aligned}\)
\(\begin{aligned} q-2017&=0\\ q&=2017\\ \sqrt{y^2-2017^2}&=2017\\ y&=2017\sqrt2 \end{aligned}\)
\(\begin{aligned} xy&=2016\sqrt2\cdot2017\sqrt2\\ &=\boxed{\boxed{8132544}} \end{aligned}\)
Jadi, nilai xy = 8132544.
JAWAB: E
JAWAB: E
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