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a, Điều kiện x ∉ {\(\frac{5}{3};\frac{1}{7}\)}
\(\sqrt{3x-5}=\sqrt{7x-1}\)
\(\left(\sqrt{3x-5}\right)^2=\left(\sqrt{7x-1}\right)^2\)
\(\left|3x-5\right|=\left|7x-1\right|\)
\(3x-5=7x-1\)
\(-4x=4\) => x = -1
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
\(a,x-3\sqrt{x}+2\)
\(=x-3\sqrt{x}+\frac{9}{4}-\frac{1}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2=\left(x+2\right)\left(x-2\right)\)
câu a mình nhìn nhầm :
\(=\left(x-1\right)\left(x+2\right)\)
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-2\\x\ge-3\\x\ge-4\\x\ge-7\end{matrix}\right.\Leftrightarrow}x\ge-2\)
\(\sqrt{x+2}-\sqrt{x+3}=\sqrt{x+4}-\sqrt{x+7}\)
\(\Leftrightarrow x+2-2\sqrt{\left(x+2\right)\left(x+3\right)}+x+3=x+4-2\sqrt{\left(x+4\right)\left(x+7\right)}+x+7\)
\(\Leftrightarrow-2\sqrt{\left(x+2\right)\left(x+3\right)}+2\sqrt{\left(x+4\right)\left(x+7\right)}=6\)
\(\Leftrightarrow2\left[\sqrt{\left(x+4\right)\left(x+7\right)}-\sqrt{\left(x+2\right)\left(x+3\right)}\right]=6\)
\(\Leftrightarrow\sqrt{\left(x+4\right)\left(x+7\right)}-\sqrt{\left(x+2\right)\left(x+3\right)}=3\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)-2\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}+\left(x+2\right)\left(x+3\right)=9\)
\(\Leftrightarrow-2\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}=-2x^2-16x-8\)
\(\Leftrightarrow\sqrt{\left(x+4\right)\left(x+7\right)\left(x+2\right)\left(x+3\right)}=x^2+8x+4\)
Có lẽ làm sai ở đâu đó, mk lười :V
ĐKXĐ: \(x\ge-2\)
\(\Leftrightarrow\sqrt{x+2}+\sqrt{x+7}=\sqrt{x+3}+\sqrt{x+4}\)
\(\Leftrightarrow2x+9+2\sqrt{x^2+9x+14}=2x+7+2\sqrt{x^2+7x+12}=0\)
\(\Leftrightarrow\sqrt{x^2+9x+14}+1=\sqrt{x^2+7x+12}\)
\(\Leftrightarrow x^2+9x+15+2\sqrt{x^2+9x+14}=x^2+7x+12\)
\(\Leftrightarrow2\sqrt{x^2+9x+14}=-2x-3\) (\(x\le-\frac{3}{2}\))
\(\Leftrightarrow4\left(x^2+9x+14\right)=4x^2+12x+9\)
\(\Leftrightarrow24x=-47\)
\(\Leftrightarrow x=-\frac{47}{24}\)