mong mọi người giải giúp em vs 

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

lamf nốt 4

a: \(2x^2-11x+21=3\cdot\sqrt[3]{4x-4}\)

=>\(2x^2-6x-5x+15=3\cdot\sqrt[3]{4x-4}-6\)

=>\(\left(x-3\right)\left(2x-5\right)=3\cdot\frac{4x-4-8}{\sqrt[3]{\left(4x-4\right)^2}+2\cdot\sqrt[3]{4x-4}+4}\)

=>\(\left(x-3\right)\left(2x-5\right)-3\cdot\frac{4x-12}{\sqrt[3]{\left(4x-4\right)^2}+2\cdot\sqrt[3]{4x-4}+4}=0\)

=>\(\left(x-3\right)\left\lbrack\left(2x-5\right)-3\cdot\frac{4}{\sqrt[3]{\left(4x-4\right)^2}+2\cdot\sqrt[3]{4x-4}+4}\right\rbrack=0\)

=>x-3=0

=>x=3

a, ĐKXĐ: \(x^2-4x+4\ge0\Rightarrow\left(x-2\right)^2\ge0\left(luônđúng\right)\)

 \(\sqrt{x^2-4x+4}=1\\ \Rightarrow x-2=1\\ \Rightarrow x=3\)

b,\(ĐKXĐ:1-4x+4x^2\ge0\Rightarrow\left(1-2x\right)^2\ge0\left(luônđúng\right)\)

 \(\sqrt{1-4x+4x^2}=5\\ \Rightarrow\left|1-2x\right|=5\\ \Rightarrow\left[{}\begin{matrix}1-2x=5\\1-2x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

d, ĐKXĐ: \(\left\{{}\begin{matrix}9x^2\ge0\\2x+1\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x\ge0\)

\(\sqrt{9x^2}=2x+1\\ \Rightarrow\left|3x\right|=2x+1\\ \Rightarrow\left[{}\begin{matrix}3x=2x+1\\3x=-2x+1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

c, ĐKXĐ: \(1-2x+x^2\ge0\Rightarrow\left(1-x\right)^2\ge0\left(luônđúng\right)\)

 \(\sqrt{1-2x+x^2}-6=0\\ \Rightarrow\left|1-x\right|=6\\ \Rightarrow\left[{}\begin{matrix}1-x=-6\\1-x=6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)

e, \(\left\{{}\begin{matrix}9-6x+x^2\ge0\\x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3-x\right)^2\ge0\left(luônđúng\right)\\x\ge0\end{matrix}\right.\)\(\Rightarrow x\ge0\)

\(\sqrt{9-6x+x^2}=x\\ \Rightarrow\left|3-x\right|=x\\ \Rightarrow\left[{}\begin{matrix}3-x=-x\\3-x=x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3=0\left(vôlí\right)\\x=1,5\end{matrix}\right.\)

c: ĐKXĐ: \(\begin{cases}x-2\ge0\\ x+2\ge0\\ x^2-4\ge0\end{cases}\)

=>x>=2 và \(x^2\ge4\)

=>x>=2

Ta có: \(\sqrt{x-2}-\sqrt{x+2}=2\cdot\sqrt{x^2-4}-2x+2\)

=>\(\sqrt{x-2}-\sqrt{x+2}+2=2\cdot\sqrt{x^2-4}-2x+4\)

=>\(\sqrt{x-2}-\frac{x+2-4}{\sqrt{x+2}+2}=2\cdot\sqrt{\left(x-2\right)\left(x+2\right)}-2\left(x-2\right)\)

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

=>\(\sqrt{x-2}\left(1-\frac{\sqrt{x-2}}{\sqrt{x+2}+2}-2\sqrt{x+2}+4\right)=0\)

=>\(\sqrt{x-2}=0\)

=>x-2=0

=>x=2(nhận)

a: ĐKXĐ: \(x^2-1\ge0\)

=>x>=1 hoặc x<=-1

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

=>\(\sqrt{x-\sqrt{x^2-1}}-1+\sqrt{x+\sqrt{x^2-1}}-1=0\)

=>\(\frac{x-\sqrt{x^2-1}-1}{\sqrt{x-\sqrt{x^2-1}+1}}+\frac{x+\sqrt{x^2-1}-1}{\sqrt{x+\sqrt{x^2-1}+1}}=0\)

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

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

=>\(\sqrt{x-1}=0\)

=>x-1=0

=>x=1(nhận)

Giải PT

a) \(3\sqrt{9x}+\sqrt{25x}-\sqrt{4x} = 3\)

\(\Leftrightarrow\) \(3.3\sqrt{x} +5\sqrt{x} - 2\sqrt{x} = 3 \)

\(\Leftrightarrow\) \(9\sqrt{x}+5\sqrt{x}-2\sqrt{x} = 3 \)

\(\Leftrightarrow\) \(12\sqrt{x} = 3\)

\(\Leftrightarrow\) \(\sqrt{x} = 4 \)

\(\Leftrightarrow\) \(\sqrt{x^2} = 4^2\)

\(\Leftrightarrow\) \(x=16\)

b) \(\sqrt{x^2-2x-1} - 3 =0\)

\(\Leftrightarrow\) \(\sqrt{(x-1)^2} -3=0\)

\(\Leftrightarrow\) \(|x-1|=3\)

* \(x-1=3\)

\(\Leftrightarrow\) \(x=4\)

* \(-x-1=3\)

\(\Leftrightarrow\) \(-x=4\)

\(\Leftrightarrow\) \(x=-4\)

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

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

<=> \(|2x+1|=3+x\)

* \(2x+1=3+x\)

<=> \(2x-x=3-1\)

<=> \(x=2\)

* \(-2x+1=3+x\)

<=> \(-2x-x = 3-1\)

<=> \(-3x=2\)

<=> \(x=\dfrac{-2}{3}\)

d) \(\sqrt{x-1} = x-3\)

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

<=> \(|x-1| = x^2-2.x.3+3^2\)

<=> \(|x-1| = x-6x+9\)

<=> \(|x-1| = -5x+9\)

* \(x-1= -5x+9\)

<=> \(x+5x = 9+1\)

<=> \(6x=10\)

<=> \(x= \dfrac{10}{6} =\dfrac{5}{3}\)

* \(-x-1 = -5x+9\)

<=> \(-x+5x = 9+1\)

<=> \(4x = 10\)

<=> \(x= \dfrac{10}{4} = \dfrac{5}{2}\)

mình nghĩ câu b \(\left(x-1\right)^2\)luôn lớn hơn 0 nên chắc không cần chia ra hai trường hợp nhỉ ?