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 $x=-6$

3.

$\sqrt{4x^2+4x+1}=6$

$\Leftrightarrow \sqrt{(2x+1)^2}=6$

$\Leftrightarrow |2x+1|=6$

$\Leftrightarrow 2x+1=6$ hoặc $2x+1=-6$

$\Leftrightarrow x=\frac{5}{2}$ hoặc $x=\frac{-7}{2}$

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

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

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

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

  \(\Rightarrow6x-1=5\)

 \(\Rightarrow6x=6\Rightarrow x=1\) 

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

  \(\Rightarrow x^2+x=x^2\)

\(\Rightarrow x=0\)   

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

\(\Rightarrow-1=0\) (vô lý)

=> PT vô nghiệm 

\(\sqrt{4x^2-20x+25}+2x=5\\ < =>\sqrt{\left(2x-5\right)^2}+2x=5\\ < =>\left|2x-5\right|+2x=5 \\ < =>\left[{}\begin{matrix}2x-5+2x=5\left(x\ge\dfrac{5}{2}\right)\\2x-5+2x=-5\left(x< \dfrac{5}{3}\right)\end{matrix}\right.< =>\left[{}\begin{matrix}4x=10< =>x=\dfrac{5}{2}\left(tmdk\right)\\4x=0< =>x=0\left(ktmdk\right)\end{matrix}\right.\\ =>x=\dfrac{5}{2}\)

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

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

\(\Leftrightarrow5-2x\ge0\) (tính chất: \(\left|A\right|=A\Leftrightarrow A\ge0\))

\(\Leftrightarrow x\le\dfrac{5}{2}\)

Vậy nghiệm của pt là \(x\le\dfrac{5}{2}\)

\(\sqrt{4x^2-20x+25}+2x=5\left(đk:x\le\dfrac{5}{2}\right)\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\2x-5=2x-5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\in R\\x\le\dfrac{5}{2}\end{matrix}\right.\)

để đâu bn

\(\sqrt{\left(2x\right)^2+2.2x.5+5^2}+\sqrt{x^2+2.x.3+3^2}=10x-20\)

\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x+3\right)^2}=10x-20\)

\(\Leftrightarrow2x+5+x+3=10x-20\)

\(\Leftrightarrow7x=28\Leftrightarrow x=4\)

so sánh \(\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)và \(\sqrt{401}\)mong ad giúp mình youtube của mình là long vh nhớ đăng ký nhé 

biểu thức trong căn biến đổi thành bp dc nên không dk nhe bạn

Căn bậc ba không cần đk nhé

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

<=>\(\sqrt{\left(2x\right)^2-2.2x.5+5^2}+2x=5\)

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

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

Xét \(2x-5\ge0\Leftrightarrow x\ge\dfrac{5}{2}\)

=>2x-5+2x=5

=>4x=10

=> x=5/2(nhận)

Xét \(2x-5\le0< =>x\le\dfrac{5}{2}\)

=>-2x+5+2x=5

=>5=5(hn)

Vậy x=5/2 hoặc x<5/2

\(\sqrt{4+20x}=3x+2\left(x\ge-\dfrac{1}{5}\right)\\ \Leftrightarrow4+20x=9x^2+12x+4\\ \Leftrightarrow9x^2-8x=0\\ \Leftrightarrow x\left(9x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=\dfrac{8}{9}\left(N\right)\end{matrix}\right.\\ \sqrt{2x+5}=x+1\left(x\ge-\dfrac{5}{2}\right)\\ \Leftrightarrow2x+5=x^2+2x+1\\ \Leftrightarrow x^2-4=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=-2\left(N\right)\end{matrix}\right.\)

\(\sqrt{4+20x}=3x+2\\ \Leftrightarrow4+20x=\left(3x+2\right)^2\\ \Leftrightarrow4+20x=9x^2+12x+4\\ \Leftrightarrow-4-20x+9x^2+12x+4=0\\ \Leftrightarrow9x^2-8x=0\\ \Leftrightarrow x\left(9x-8\right)=0\\ \Leftrightarrow x=0hoặcx=\dfrac{8}{9}\)

\(\sqrt{2x+5}=x+1\\ \Leftrightarrow2x+5=\left(x+1\right)^2\\ \Leftrightarrow2x+5=x^2+2x+1\\ \Leftrightarrow x^2+2x+1-2x-5=0\\ \Leftrightarrow x^2-4=0\\ \Leftrightarrow x^2=4\\ \Leftrightarrow x=\pm2\)

a: Ta có: \(\sqrt{20x+4}=3x+2\)

\(\Leftrightarrow9x^2+12x+4=20x+4\)

\(\Leftrightarrow9x^2-8x=0\)

\(\Leftrightarrow x\left(9x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\dfrac{8}{9}\left(nhận\right)\end{matrix}\right.\)