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No.
Diketahui $x,y\in R$, $x\gt2016$ dan $x\gt2017$. Jika $2016\sqrt{(x+2016)(x-2016)}+2017\sqrt{(x+2017)(x-2017)}=\dfrac12\left(x^2+y^2\right)$maka nilai $xy=$
- $4066272$
- $4068289$
- $5750577{,}011$
- $5756281{,}$
- $8132544$
ALTERNATIF PENYELESAIAN
Misal $a=\sqrt{(x+2016)(x-2016)}$
\begin{aligned} a^2&=x^2-2016^2\\ x^2&=a^2+2016^2 \end{aligned} Misal $b=\sqrt{(y+2017)(y-2017)}$
\begin{aligned} b^2&=y^2-2017^2\\ y^2&=b^2+2017^2 \end{aligned} \begin{aligned} 2016a+2017b&=\dfrac12\left(a^2+2016^2+b^2+2017^2\right)\\ 2\cdot2016a+2\cdot2017b&=a^2+2016^2+b^2+2017^2\\ a^2-2\cdot2016a+2016^2+b^2-2\cdot2017b+2017^2&=0\\ (a-2016)^2+(b-2017)^2&=0 \end{aligned} didapat $a=2016$ dan $b=2017$
\begin{aligned} x^2&=2016^2+2016^2\\ &=2\cdot2016^2\\ x&=2016\sqrt2 \end{aligned} \begin{aligned} y^2&=2017^2+2017^2\\ &=2\cdot2017^2\\ x&=2017\sqrt2 \end{aligned} \begin{aligned} xy&=2016\sqrt2\cdot2017\sqrt2\\ &=\boxed{\boxed{8132544}} \end{aligned}
\begin{aligned} a^2&=x^2-2016^2\\ x^2&=a^2+2016^2 \end{aligned} Misal $b=\sqrt{(y+2017)(y-2017)}$
\begin{aligned} b^2&=y^2-2017^2\\ y^2&=b^2+2017^2 \end{aligned} \begin{aligned} 2016a+2017b&=\dfrac12\left(a^2+2016^2+b^2+2017^2\right)\\ 2\cdot2016a+2\cdot2017b&=a^2+2016^2+b^2+2017^2\\ a^2-2\cdot2016a+2016^2+b^2-2\cdot2017b+2017^2&=0\\ (a-2016)^2+(b-2017)^2&=0 \end{aligned} didapat $a=2016$ dan $b=2017$
\begin{aligned} x^2&=2016^2+2016^2\\ &=2\cdot2016^2\\ x&=2016\sqrt2 \end{aligned} \begin{aligned} y^2&=2017^2+2017^2\\ &=2\cdot2017^2\\ x&=2017\sqrt2 \end{aligned} \begin{aligned} xy&=2016\sqrt2\cdot2017\sqrt2\\ &=\boxed{\boxed{8132544}} \end{aligned}
No.
Jika ${n-\dfrac1n=x}$, maka berapakah ${n^2+\dfrac1{n^2}}$ jika dinyatakan dalam $x$?- ${x^2+1}$
- ${x+1}$
- ${x^3+1}$
- ${x^2+2}$
- ${\sqrt{x}+\sqrt1}$
ALTERNATIF PENYELESAIAN
\(\eqalign{
n-\dfrac1n&=x\\
\left(n-\dfrac1n\right)^2&=x^2\\
n^2-2(n)\left(\dfrac1n\right)+\left(\dfrac1n\right)^2&=x^2\\
n^2-2+\dfrac1{n^2}&=x^2\\
n^2+\dfrac1{n^2}&=\boxed{\boxed{x^2+2}}
}\)
No.
Jika ${\dfrac{3a+4b}{2a-2b}=5}$ maka tentukan nilai dari $\dfrac{a^2+2b^2}{ab}$ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{3a+4b}{2a-2b}&=5\\
3a+4b&=10a-10b\\
7a&=14b\\
\dfrac{a}b&=2
\end{aligned}
\begin{aligned}
\dfrac{a^2+2b^2}{ab}&=\dfrac{a}b+2\dfrac{b}a\\
&=2+2\left(\dfrac12\right)\\
&=\boxed{\boxed{3}}
\end{aligned}
No.
Nilai dari $\dfrac{(1945+2011)^2+(2011-1945)^2}{1945^2+2011^2}$ adalah ....ALTERNATIF PENYELESAIAN
Misal $a=1945$ dan $b=2011$
\begin{aligned} \dfrac{(1945+2011)^2+(2011-1945)^2}{1945^2+2011^2}&=\dfrac{(a+b)^2+(b-a)^2}{a^2+b^2}\\ &=\dfrac{a^2+2ab+b^2+b^2-2ab+a^2}{a^2+b^2}\\ &=\dfrac{2\left(a^2+b^2\right)}{a^2+b^2}\\ &=\boxed{\boxed{2}} \end{aligned}
\begin{aligned} \dfrac{(1945+2011)^2+(2011-1945)^2}{1945^2+2011^2}&=\dfrac{(a+b)^2+(b-a)^2}{a^2+b^2}\\ &=\dfrac{a^2+2ab+b^2+b^2-2ab+a^2}{a^2+b^2}\\ &=\dfrac{2\left(a^2+b^2\right)}{a^2+b^2}\\ &=\boxed{\boxed{2}} \end{aligned}
No.
Bilangan-bilangan real $a$, $b$, dan $c$ memenuhi sistem persamaan ${a+b=8}$ dan ${ab=c^2+16}$. Hasil dari ${a+b+c=}$ ....- $4$
- $5$
- $6$
- $7$
- $8$
ALTERNATIF PENYELESAIAN
\begin{aligned}
ab&\leq\dfrac{(a+b)^2}4\\
&\leq\dfrac{8^2}4\\
&\leq16
\end{aligned}
\begin{aligned}
c^2+16&\geq16\\
ab&\geq16\\
\end{aligned}
Didapat ${ab=16}$ dan ${c=0}$
$a+b+c=8+0=\boxed{\boxed{8}}$
$a+b+c=8+0=\boxed{\boxed{8}}$
No.
Jika $x+\dfrac1x=4$ maka nilai $x^3+x^{-3}=$ ....ALTERNATIF PEMBAHASAN
$\eqalign{
\left(x+\dfrac1x\right)^2&=4^2\\
x^2+2+\dfrac1{x^2}&=16\\
x^2+\dfrac1{x^2}&=14
}$
$\eqalign{ \left(x^2+\dfrac1{x^2}\right)\left(x+\dfrac1x\right)&=(14)(4)\\ x^3+x+\dfrac1x+\dfrac1{x^3}&=56\\ x^3+4+\dfrac1{x^3}&=56\\ x^3+\dfrac1{x^3}&=\boxed{\boxed{52}} }$
$\eqalign{ \left(x^2+\dfrac1{x^2}\right)\left(x+\dfrac1x\right)&=(14)(4)\\ x^3+x+\dfrac1x+\dfrac1{x^3}&=56\\ x^3+4+\dfrac1{x^3}&=56\\ x^3+\dfrac1{x^3}&=\boxed{\boxed{52}} }$
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