a: ĐKXĐ: 2x+5>=0 và 1-x>=0

=>-5/2<=x<=1

PT =>2x+5=1-x

=>3x=-4

=>x=-4/3(nhận)

b: ĐKXĐ: x^2-x>=0 và 3-x>=0

=>x<=3 và (x>=1 hoặc x<=0)

=>x<=0 hoặc (1<=x<=3)

PT =>x^2-x=3-x

=>x^2=3

=>x=căn 3(nhận) hoặc x=-căn 3(nhận)

c: ĐKXĐ: 2x^2-3>=0 và 4x-3>=0

=>x>=3/4 và x^2>=3/2

=>x>=3/4 và \(\left[{}\begin{matrix}x>=\dfrac{\sqrt{6}}{4}\\x< =\dfrac{-\sqrt{6}}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x>=\dfrac{3}{4}\\x< =-\dfrac{\sqrt{6}}{4}\end{matrix}\right.\)

PT =>2x^2-3=4x-3

=>2x^2-4x=0

=>2x(x-2)=0

=>x=0(loại) hoặc x=2(nhận)

\(\sqrt{2x+5}=\sqrt{1-x}\) (ĐK: \(-\dfrac{5}{2}\le x\le1\)) 

\(\Leftrightarrow2x+5=1-x\)

\(\Leftrightarrow2x+x=1-5\)

\(\Leftrightarrow3x=-4\)

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

b) \(\sqrt{x^2-x}=\sqrt{3-x}\) (ĐK: \(\left[{}\begin{matrix}1\le x\le3\\x\le0\end{matrix}\right.\))

\(\Leftrightarrow x^2-x=3-x\)

\(\Leftrightarrow x^2=3\)  

\(\Leftrightarrow x=\pm\sqrt{3}\left(tm\right)\)  

c) \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (ĐK: \(x\ge\dfrac{\sqrt{6}}{2}\)) 

\(\Leftrightarrow2x^2-3=4x-3\)

\(\Leftrightarrow2x^2=4x\)

\(\Leftrightarrow x^2=2x\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

a: Ta có: \(\sqrt{\left(x-3\right)^2}=3-x\)

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

\(\Leftrightarrow x-3\le0\)

hay \(x\le3\)

b: Ta có: \(\sqrt{4x^2-20x+25}+2x=5\)

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

\(\Leftrightarrow2x-5\le0\)

hay \(x\le\dfrac{5}{2}\)

1.

$\sqrt{3x^2}-\sqrt{12}=0$

$\Leftrightarrow \sqrt{3x^2}=\sqrt{12}$

$\Leftrightarrow 3x^2=12$

$\Leftrightarrow x^2=4$

$\Leftrightarrow (x-2)(x+2)=0\Leftrightarrow x=\pm 2$

2. 

$\sqrt{(x-3)^2}=9$

$\Leftrightarrow |x-3|=9$

$\Leftrightarrow x-3=9$ hoặc $x-3=-9$

$\Leftrightarrow x=12$ hoặc $x=-6$

Nguyễn Thành Trương , mình đang sài latop, nhìn bài của cậu, tớ muốn quẹo cả cổ -.-

Hoài Dung Copy ảnh. Mở paint past vào chỉnh hướng rồi xem :)

Giải:

a) \(\sqrt{\left(x-3\right)^2}=3-x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x=6\\0x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\0x=0\end{matrix}\right.\)

Vậy ...

b) \(\sqrt{25-20x+4x^2}+2x=5\)

\(\Leftrightarrow\sqrt{5^2-2.5.2x+\left(2x\right)^2}+2x=5\)

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\-4x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy ...

c) \(\sqrt{1-12x+36x^2}=5\)

\(\Leftrightarrow\sqrt{\left(1-6x\right)^2}=5\)

\(\Leftrightarrow\left|1-6x\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}1-6x=5\\1-6x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=1-5\\6x=1-\left(-5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}6x=-4\\6x=6\end{matrix}\right.\)

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

Vậy ...

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

\(a,\sqrt{4x^2-20x+25}+2x=5\)

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

  \(\Rightarrow4x=10\Rightarrow x=\frac{5}{2}\)

\(b,\sqrt{1-12x+36x^2}=5\)

  \(\Rightarrow6x-1=5\)

 \(\Rightarrow6x=6\Rightarrow x=1\) 

\(c,\sqrt{x^2+x}=x\)

  \(\Rightarrow x^2+x=x^2\)

\(\Rightarrow x=0\)   

\(c,\Rightarrow\left(x-2\right)^2-1=\left(x-2\right)^2\)

\(\Rightarrow-1=0\) (vô lý)

=> PT vô nghiệm 

a: ĐKXĐ: x>=1

\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)

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

=>-2*căn x-1=-2

=>căn x-1=1

=>x-1=1

=>x=2

b: ĐKXĐ: x>=1

\(PT\Leftrightarrow\sqrt{x-1}\cdot\dfrac{1}{2}-\dfrac{9}{2}\cdot\sqrt{x-1}+\dfrac{24\sqrt{x-1}}{8}=-17\)

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

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

=>x-1=289

=>x=290

a)\(ĐKXĐ:x\ge\frac{-1}{2}\)

 \(\sqrt{x^2+4x+4}=2x+1\)

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

\(\Leftrightarrow x+2=2x+1\)

\(\Leftrightarrow-x=-1\)

\(\Leftrightarrow x=1\)

Vậy nghiệm duy nhất của phương trình là 1.

b)\(ĐKXĐ:x\ge3\)

 \(\sqrt{4x^2-12x+9}=x-3\)

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

\(\Leftrightarrow2x-3=x-3\)

\(\Leftrightarrow2x=x\)

\(\Leftrightarrow x=0\)(không t/m đkxđ)

Vậy phương trình vô nghiệm