\(ĐK:x>-\dfrac{5}{7}\\ PT\Leftrightarrow7x+5=9x-7\Leftrightarrow x=6\left(tm\right)\)

a/ \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐKXĐ : \(x\ge1\))

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow2\sqrt{x-1}=2\Leftrightarrow x-1=1\Leftrightarrow x=2\)

b/ \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)

\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}+3=0\)

<=> 3 = 0 (vô lý)

=> pt vô nghiệm.

c/ \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (ĐKXĐ : x>-5/7)

\(\Leftrightarrow9x-7=7x+5\Leftrightarrow2x=12\Leftrightarrow x=6\)

d/ \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\) (ĐKXĐ : \(x\ge\frac{3}{2}\))

\(\Leftrightarrow2x-3=4\left(x-1\Leftrightarrow\right)2x=1\Leftrightarrow x=\frac{1}{2}\) (loại)

Vậy pt vô nghiệm.

\(ĐK:x>-\dfrac{5}{7}\\ PT\Leftrightarrow9x-7=7x+5\\ \Leftrightarrow2x=12\Leftrightarrow x=6\left(tm\right)\)

ĐK: \(x>\frac{-5}{7}\)

<=> 9x-7=7x+5 <=> x=6 (thõa mãn)

a/ ĐKXĐ: \(x\ge\frac{-5}{7}\)

\(\Leftrightarrow9x-7=7x+5\Leftrightarrow x=6\)(thoả mãn)

b/ ĐKXĐ:....

\(\Leftrightarrow2x^2-3=4x-3\Leftrightarrow\left[{}\begin{matrix}x=2\left(thoảman\right)\\x=0\left(loai\right)\end{matrix}\right.\)

c/ ĐKXĐ:...

\(\Leftrightarrow\frac{2x-3}{x-1}=4\Leftrightarrow2x-3=4x-4\Leftrightarrow x=\frac{1}{2}\)(thoả mãn)

d/ giống câu c nhưng đkxđ khác và nó vô no

a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (1)

\(\Leftrightarrow9x-7=\sqrt{\left(7x+5\right)\left(7x+5\right)}\)

\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)\left(7x+5\right)}=7\)

\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)^2}=7\)

\(\Leftrightarrow9x-\left|7x+5\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}9x-\left(7x+5\right)=7\left(đk:7x+5\ge0\right)\\9x-\left[-\left(7x+5\right)\right]=7\left(đk:7x+5< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(đk:x\ge-\dfrac{5}{7}\right)\\x=\dfrac{1}{8}\left(đk:x< -\dfrac{5}{7}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x\in\varnothing\end{matrix}\right.\)

\(\Leftrightarrow x=6\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{6\right\}\)

b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\) (2)

\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3\cdot\dfrac{\sqrt{x+5}}{3}-\dfrac{1}{3}\cdot\sqrt{9\left(x-5\right)}=4\)

\(\Leftrightarrow\sqrt{4}\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot\sqrt{9}\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}+\sqrt{x+5}=4\)

\(\Leftrightarrow\sqrt{x-5}=4-\sqrt{x+5}\)

\(\Leftrightarrow x-5=\left(4-\sqrt{x+5}\right)^2\)

\(\Leftrightarrow x-5=16-8\sqrt{x+5}+x+5\)

\(\Leftrightarrow-5=16-8\sqrt{x+5}+5\)

\(\Leftrightarrow-5=21-8\sqrt{x+5}\)

\(\Leftrightarrow8\sqrt{x+5}=21+5\)

\(\Leftrightarrow8\sqrt{x+5}=26\)

\(\Leftrightarrow\sqrt{x+5}=\dfrac{13}{4}\)

\(\Leftrightarrow x+5=\dfrac{169}{16}\)

\(\Leftrightarrow x=\dfrac{169}{16}-5\)

\(\Leftrightarrow x=\dfrac{89}{16}\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{89}{16}\right\}\)

\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow9x-7x=5+7\)

\(\Leftrightarrow2x=12\)

\(\Leftrightarrow x=6\)

\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\)

\(\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\)

\(\Leftrightarrow x=9\)