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a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)
=>4x-4=2x-3
=>2x=1
hay x=1/2
b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>(2x-3)=4x-4
=>4x-4=2x-3
=>2x=1
hay x=1/2(nhận)
c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=-3/2 hoặc x=7/2
e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
=>căn (x-5)=2
=>x-5=4
hay x=9
\(A=3\sqrt{8}-\sqrt{50}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow6\sqrt{2}-5\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(\Leftrightarrow\sqrt{2}-\sqrt{\sqrt{2}-1}\)
\(B=2.\dfrac{2}{x-1}.\sqrt{\dfrac{x^2-2x+1}{4x^2}}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{x^2-2x+1}}{2x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{\sqrt{\left(x-1\right)^2}}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x-1}.\dfrac{x-1}{x}\)
\(\Leftrightarrow\)\(2.\dfrac{1}{x}\)
\(\Leftrightarrow\)\(\dfrac{2}{x}\)
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
Bài 3:
a: \(=\left(4\sqrt{2}-6\sqrt{2}\right)\cdot\dfrac{\sqrt{2}}{2}=-2\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=-2\)
b: \(=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-2\left(\sqrt{6}-1\right)\)
\(=\sqrt{6}-2\sqrt{6}+2=2-\sqrt{6}\)
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