a) \(\sqrt[]{x^2-4x+4}=x+3\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

2: =>2x^2-8x+4=x^2-4x+4 và x>=2

=>x^2-4x=0 và x>=2

=>x=4

3: \(\sqrt{x^2+x-12}=8-x\)

=>x<=8 và x^2+x-12=x^2-16x+64

=>x<=8 và x-12=-16x+64

=>17x=76 và x<=8

=>x=76/17

4: \(\sqrt{x^2-3x-2}=\sqrt{x-3}\)

=>x^2-3x-2=x-3 và x>=3

=>x^2-4x+1=0 và x>=3

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

6:

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

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

=>\(\left|\sqrt{x-1}-1\right|=\sqrt{x-1}+1+2=\sqrt{x-1}+3\)

=>1-căn x-1=căn x-1+3 hoặc căn x-1-1=căn x-1+3(loại)

=>-2*căn x-1=2

=>căn x-1=-1(loại)

=>PTVN

1) ĐK: \(x\ge\dfrac{5}{2}\)

pt <=> \(x-4=\sqrt{2x-5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left(x-4\right)^2=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-8x+16=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-10x+21=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left(x-3\right)\left(x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\\left[{}\begin{matrix}x=3\left(l\right)\\x=7\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=7

2) ĐK: \(2x^2-8x+4\ge0\)

pt <=> \(\left\{{}\begin{matrix}x\ge2\\2x^2-8x+4=x^2-4x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\left(x-4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=0\left(l\right)\\x=4\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy, pt có nghiệm duy nhất là x=4

3) ĐK: \(x\ge3\)

pt <=> \(\left\{{}\begin{matrix}x\le8\\x^2+x-12=x^2-16x+64\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\17x=76\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x=\dfrac{76}{17}\left(n\right)\end{matrix}\right.\) 

Vậy, pt có nghiệm duy nhất là \(x=\dfrac{76}{17}\)\(\)

a/ \(2x-3=5x+2\)

\(\Leftrightarrow5x-2x=-3-2\)

\(\Leftrightarrow3x=-5\Leftrightarrow x=-\dfrac{5}{3}\)

Vậy..

b. \(2x\left(x-1\right)=2x+2\)

\(\Leftrightarrow2x^2-4x-2=0\)

\(\Leftrightarrow x^2-2x-1=0\)

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

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

Vậy...

c/ ĐKXĐ : \(x\ne\pm2\)

\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{6}{\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow x^2+4x+4-x^2=6x-12\)

\(\Leftrightarrow2x-16=0\)

\(\Leftrightarrow x=8\)

Vậy..

Phần b bằng bn vậy ? 

hết cứu đi mà làm

giải phaj bỏ ngoặc nhức đầu lắm

Đặt \(2x^2-2x+2=a\)

\(\Leftrightarrow\dfrac{a-3x}{x-1}=\dfrac{a+3x+15}{x-3}\)

\(\Leftrightarrow6x^2+3x+2a-15=0\)

\(\Leftrightarrow10x^2-x-11=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=-1\end{matrix}\right.\)

E chưa hiểu cách này lắm ạ.