a)ĐK:\(\begin{cases}25x^2-9 \ge 0\\5x+3 \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}(5x-3)(5x+3) \ge 0\\5x+3 \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge \dfrac35\\x \le -\dfrac35\end{array} \right.\\\end{cases}\)

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x \ge \dfrac35\end{array} \right.\)

`pt<=>\sqrt{5x+3}(\sqrt{5x-3}-2)=0`

`<=>` \(\left[ \begin{array}{l}5x+3=0\\\sqrt{5x-3}=2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\5x-3=4\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x=7/5\end{array} \right.\) 

`b)sqrt{x-3}/sqrt{2x+1}=2`

ĐK:\(\begin{cases}x-3 \ge 0\\2x+1>0\\\end{cases}\)

`<=>x>=3`

`pt<=>sqrt{x-3}=2sqrt{2x+1}`

`<=>x-3=8x+4`

`<=>7x=7`

`<=>x=1(l)`

`c)sqrt{x^2-2x+1}+sqrt{x^2-4x+4}=3`

`<=>sqrt{(x-1)^2}+sqrt{(x-2)^2}=3`

`<=>|x-1|+|x-2|=3`

`**x>=2`

`pt<=>x-1+x-2=3`

`<=>2x=6`

`<=>x=3(tm)`

`**x<=1`

`pt<=>1-x+2-x=3`

`<=>3-x=3`

`<=>x=0(tm)`

`**1<=x<=2`

`pt<=>x-1+2-x=3`

`<=>=-1=3` vô lý

Vậy `S={0,3}`

\(\sqrt{25x^2-10x+1}=4x+9\)

\(\Leftrightarrow\sqrt{\left(5x-1\right)^2}=4x+9\)

\(\Leftrightarrow\left|5x-1\right|=4x+9\)

\(\Leftrightarrow\orbr{\begin{cases}5x-1=4x+9\\5x-1=-4x-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=10\\x=-\frac{8}{9}\end{cases}}}\)

Vậy ... 

\(\sqrt{x^2+2x+1}=\sqrt{x+1}\)

\(\Leftrightarrow\sqrt{\left(x+1\right)^2}=\sqrt{x+1}\)

\(\Leftrightarrow\sqrt{\left(x+1\right)^2}-\sqrt{x+1}=0\)

\(\Leftrightarrow\sqrt{x+1}.\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x+1}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)

Vậy ...

\(\sqrt{x^2+2x+1}=\sqrt{x+1}\)

\(\Rightarrow\sqrt{\left(x+1\right)^2}=\sqrt{x+1}\)

\(\Rightarrow x+1=\sqrt{x+1}\)

\(\Rightarrow\left(x+1\right)^2=x+1\)

\(\Rightarrow x^2+2x+1=x+1\)

\(\Rightarrow x^2+x=0\)

\(\Rightarrow x\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

a) ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow x-2=x^2+2x+1\)

\(\Leftrightarrow x^2+x+3=0\)(vô lý do \(x^2+x+3=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\))

Vậy \(S=\varnothing\)

b) ĐKXĐ: \(x\ge-3\)

\(pt\Leftrightarrow1+x^2=x^2+6x+9\)

\(\Leftrightarrow6x=-8\Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)

a) giải pt ra ta được  : x=-1

b) giải pt ra ta được  : x=2

c)giải pt ra ta được  : x vô ngiệm

d)giải pt ra ta được  : x=vô ngiệm

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

a) \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

<=> \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)

<=>\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

<=>\(\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+\frac{6}{2}\right)=-17\)

<=>\(\sqrt{x-1}=-17\)

<=>x-1=17

<=>x=18

Vậy pt có nghiệm là x=18

\(a.ĐK:x-1\ge0\Leftrightarrow x\ge1\)

\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{27}{2}\sqrt{x-1}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{27}{2}+24\sqrt{\frac{1}{64}}\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}.\left(-10\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=\frac{-17}{-10}=\frac{17}{10}\)

\(\Leftrightarrow x-1=\left(\frac{17}{10}\right)^2\)

\(\Leftrightarrow x=\frac{289}{100}+1=3,89\left(TM\right)\)

Vậy \(S=\left\{3,89\right\}\)

\(b.ĐK:x^2+2\ge0\)

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

\(\Leftrightarrow9\sqrt{x^2+2}+2\sqrt{x^2+2}-25\sqrt{x^2+2}=-3\)

\(\Leftrightarrow\sqrt{x^2+2}\left(9+2-25\right)=-3\)

\(\Leftrightarrow\sqrt{x^2+2}=\frac{-3}{-14}=\frac{3}{14}\)

\(\Leftrightarrow x^2+2=\left(\frac{3}{14}\right)^2\)

\(\Leftrightarrow x=\sqrt{\frac{9}{196}-2}=\sqrt{-\frac{383}{196}}\left(vl\right)\)

Vậy \(S=\varnothing\)

Mấy câu kia làm tương tự

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

<=>\(\sqrt{9\left(x^2+2\right)}+2\sqrt{x^2+2}-\sqrt{25\left(x^2+2\right)}+3=0\)

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

<=>\(\sqrt{x^2+2}\left(3+2-5\right)=-3\)

<=>0x=-3

Vậy pt vô nghiệm

a) \(\Leftrightarrow\sqrt{\left(x+3\right)^2}=4\)

\(\Leftrightarrow\left|x+3\right|=4\) \(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) ( TM )

b) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5x+3\)

\(\Leftrightarrow\left|2x-1\right|=5x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+3\ge0\\\left[{}\begin{matrix}2x-1=5x+3\\2x-1=-5x-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{3}{5}\\\left[{}\begin{matrix}x=-\frac{4}{3}\left(KTM\right)\\x=-\frac{2}{7}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)

a \(\sqrt{x^2+6x+9}=4\Leftrightarrow\sqrt{\left(x+3\right)^2=4}\)

\(\Leftrightarrow x+3=4\)

\(\Rightarrow x=1\)

a) \(\Leftrightarrow\sqrt{3}\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\sqrt{3}-1\right)=0\Leftrightarrow x=1\)

b) \(\Leftrightarrow\sqrt{\left(x-3\right)^2}=7\)

\(\Leftrightarrow\left|x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\)

c) \(\Leftrightarrow3\left|x-2\right|=45\)

\(\Leftrightarrow\left|x-2\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=15\\x-2=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-13\end{matrix}\right.\)

\(a,PT\Leftrightarrow\sqrt{3}\left(x-1\right)=1-x\\ \Leftrightarrow\sqrt{3}\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(\sqrt{3}+1\right)=0\\ \Leftrightarrow x=1\left(\sqrt{3}+1\ne0\right)\\ b,ĐK:x\in R\\ PT\Leftrightarrow\left|x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}x-3=7\\3-x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\\ c,ĐK:x\in R\\ PT\Leftrightarrow3\left|x-2\right|=45\Leftrightarrow\left|x-2\right|=15\\ \Leftrightarrow\left[{}\begin{matrix}x-2=15\\2-x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=17\\x=-13\end{matrix}\right.\)