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}
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}
No.
$\dfrac{\sqrt{3^{2015}}}{\sqrt{3^{2015}}-\sqrt{3^{2013}}}=$- $\dfrac32$
- $\dfrac43$
- $\dfrac54$
- $\dfrac65$
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}
No.
Jika $a$ dan $b$ bilangan asli dan ${\sqrt{18+\sqrt{308}}=\sqrt{a}+\sqrt{b}}$, maka nilai $a\times b$ 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}
$a\times b=77$
$a\times b=77$
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 $n\leq\dfrac{625}4$
\(\eqalign{ x&=\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}}+\sqrt{\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\\ &=\sqrt{\left(\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}}+\sqrt{\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\right)^2}\\ &=\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}+2\sqrt{\left(\dfrac{25}2+\sqrt{\dfrac{625}4-n}\right)\left(\dfrac{25}2-\sqrt{\dfrac{625}4-n}\right)}+\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\\ &=\sqrt{25+2\sqrt{\dfrac{625}4-\dfrac{625}4+n}}\\ &=\sqrt{25+2\sqrt{n}}\geq\sqrt{25}=5 }\)
$n=\left(\dfrac{x^2-25}2\right)^2$
Jika $x$ genap maka $x^2-25$ ganjil, sehingga $n$ bukan bilangan bulat. Jadi $x$ haruslah ganjil. Agar $n\leq\dfrac{625}4$ maka maksimal $x^2$ adalah $50$. Sehingga kita dapat $x=5$ dan $x=7$.
Untuk $x=5$, $n=0$
Untuk $x=7$, $n=144$
\(\eqalign{ x&=\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}}+\sqrt{\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\\ &=\sqrt{\left(\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}}+\sqrt{\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\right)^2}\\ &=\sqrt{\dfrac{25}2+\sqrt{\dfrac{625}4-n}+2\sqrt{\left(\dfrac{25}2+\sqrt{\dfrac{625}4-n}\right)\left(\dfrac{25}2-\sqrt{\dfrac{625}4-n}\right)}+\dfrac{25}2-\sqrt{\dfrac{625}4-n}}\\ &=\sqrt{25+2\sqrt{\dfrac{625}4-\dfrac{625}4+n}}\\ &=\sqrt{25+2\sqrt{n}}\geq\sqrt{25}=5 }\)
$n=\left(\dfrac{x^2-25}2\right)^2$
Jika $x$ genap maka $x^2-25$ ganjil, sehingga $n$ bukan bilangan bulat. Jadi $x$ haruslah ganjil. Agar $n\leq\dfrac{625}4$ maka maksimal $x^2$ adalah $50$. Sehingga kita dapat $x=5$ dan $x=7$.
Untuk $x=5$, $n=0$
Untuk $x=7$, $n=144$
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}
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}
$\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}
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}
0 Komentar
Silahkan berkomentar dengan santun di sini. Anda juga boleh bertanya soal matematika atau mengoreksi jawaban di atas