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a) \(2\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=28\) (*)
đk: x >/ 0
(*) \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)
\(\Leftrightarrow13\sqrt{2x}=28\) \(\Leftrightarrow\sqrt{2x}=\dfrac{28}{13}\Leftrightarrow2x=\left(\dfrac{28}{13}\right)^2\Leftrightarrow x=\dfrac{392}{169}\left(N\right)\)
Kl: \(x=\dfrac{392}{169}\)
b) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\) (*)
đk: x >/ 5
(*) \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\Leftrightarrow x-5=4\Leftrightarrow x=9\left(N\right)\)
Kl: x=9
c) \(\sqrt{\dfrac{3x-2}{x+1}}=2\) (*)
Đk: \(\left[{}\begin{matrix}x< -1\\x\ge\dfrac{2}{3}\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{3x-2}{x+1}=4\Leftrightarrow3x-2=4x+4\Leftrightarrow x=-6\left(N\right)\)
Kl: x=-6
d) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (*)
Đk: \(x\ge\dfrac{4}{5}\)
(*) \(\Leftrightarrow\sqrt{5x-4}=2\sqrt{x+2}\Leftrightarrow5x-4=4x+8\Leftrightarrow x=12\left(N\right)\)
Kl: x=12
\(a,=5\sqrt{5}-12\sqrt{5}+6\sqrt{5}-4\sqrt{5}=-5\sqrt{5}\)
\(\sqrt{29^2-20^2}=\sqrt{\left(29-20\right)\left(29+20\right)}=\sqrt{3^2.7^2}=21\)
\(\text{Đặt: }\)\(\hept{\begin{cases}\sqrt{4-\sqrt{15}}=a\\\sqrt{4+\sqrt{15}}=b\end{cases}}\)\(\text{cần tính: a-b}\)
\(\hept{\begin{cases}ab=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=1\\a^2+b^2=8\end{cases}}\Rightarrow\left(a-b\right)^2=6\Rightarrow a-b=-\sqrt{6}\left(vì:a< b\right)\)
a) Ta có: \(\sqrt{4x-8}+5\sqrt{x-2}-\sqrt{9x-18}=20\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow\sqrt{4}.\sqrt{x-2}+5\sqrt{x-2}-\sqrt{9}.\sqrt{x-2}=20\)
\(\Leftrightarrow2.\sqrt{x-2}+5\sqrt{x-2}-3.\sqrt{x-2}=20\)
\(\Leftrightarrow4.\sqrt{x-2}=20\)
\(\Leftrightarrow\sqrt{x-2}=5\)
\(\Leftrightarrow x-2=25\)
\(\Leftrightarrow x=27\left(TM\right)\)
Vậy \(S=\left\{27\right\}\)
mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)
1.
\(DK:x\in\left[-4;5\right]\)
\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)
Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)
\(\Rightarrow\sqrt{x-5}=0\)
\(x=5\left(n\right)\)
Vay nghiem cua PT la \(x=5\)
2.
\(DK:x\ge0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\)
\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)
Ta co:
\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)
Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)
TH1:
\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)
TH2:(loai)
Vay nghiem cua PT la \(x\in\left[4;9\right]\)
Lời giải:
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 2\sqrt{2x}-10\sqrt{2x}+5\sqrt{x}=-20$
$\Leftrightarrow 5\sqrt{x}-8\sqrt{2x}=-20$
$\Leftrightarrow \sqrt{x}(5-8\sqrt{2})=-20$
$\Leftrightarrow \sqrt{x}=\frac{20}{8\sqrt{2}-5}$
$\Rightarrow x=(\frac{20}{8\sqrt{2}-5})^2$
b. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{5x}-5\sqrt{3x}+4\sqrt{x}=10$
$\Leftrightarrow \sqrt{x}(3\sqrt{5}-5\sqrt{3}+4)=10$
$\Leftrightarrow \sqrt{x}=\frac{10}{3\sqrt{5}-5\sqrt{3}+4}$
$\Rightarrow x=(\frac{10}{3\sqrt{5}-5\sqrt{3}+4})^2$