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

\(\Leftrightarrow\left|3-2x\right|=4+x\)

\(\Leftrightarrow\orbr{\begin{cases}3-2x=4+x\\3-2x=-4-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=-1\\x=7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=\sqrt{7}\\x=-\sqrt{7}\end{cases}}\\\left(x-3\right)\left(x-1\right)=0\end{cases}}\)

3: Ta có: \(\sqrt{4x+1}=x+1\)

\(\Leftrightarrow x^2+2x+1=4x+1\)

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

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

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vd1: 

d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)

\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)

\(\Leftrightarrow x=6\)

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

??

Đề kiểu gì vậy ?

a, ĐKXĐ: \(x\le2\)

\(\sqrt{4-2x}=5\\ \Leftrightarrow4-2x=25\\ \Leftrightarrow2x=-21\\ \Leftrightarrow x=-10,5\left(tm\right)\)

b, ĐKXĐ: \(x\ge-1\)

\(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\\ \Leftrightarrow5\sqrt{x+1}+\sqrt{9\left(x+1\right)}=16\\ \Leftrightarrow5\sqrt{x+1}+3\sqrt{x+1}=16\\ \Leftrightarrow8\sqrt{x+1}=16\\ \Leftrightarrow\sqrt{x+1}=2\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\)

c, \(\sqrt{4x^2+12x+9}=4\Leftrightarrow4x^2+12x+9=16\\ \Leftrightarrow4x^2+12x-7=0\\ \Leftrightarrow\left(4x^2-2x\right)+\left(14x-7\right)=0\\ \Leftrightarrow2x\left(2x-1\right)+7\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

a: \(\Leftrightarrow4-2x=25\)

hay \(x=-\dfrac{21}{2}\)

c: \(\Leftrightarrow\left|2x+3\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4\\2x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

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

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

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

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

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a: ĐKXĐ: x>=2

Ta có: \(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\frac{x-2}{4}}=26\)

=>\(4\sqrt{x-2}+3\sqrt{x-2}-\frac12\cdot\sqrt{x-2}=26\)

=>\(6,5\cdot\sqrt{x-2}=26\)

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

=>x-2=16

=>x=18(nhận)

b: ĐKXĐ: x∈R

\(3x+\sqrt{4x^2-8x+4}=1\)

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

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

=>3x-1+|2x-2|=0(1)

TH1: x>=1

(1) sẽ trở thành: 3x-1+2x-2=0

=>5x-3=0

=>5x=3

=>x=3/5(loại)

TH2: x<1

(1) sẽ trở thành: 3x-1-2x+2=0

=>x+1=0

=>x=-1(nhận)

c: ĐKXĐ: x>=0

Ta có: \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\)

=>\(2x-4\sqrt{x}+\sqrt{x}-2-7=0\)

=>\(2x-3\sqrt{x}-9=0\)

=>\(2x-6\sqrt{x}+3\sqrt{x}-9=0\)

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

=>\(\sqrt{x}-3=0\)

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

=>x=9(nhận)

a,ĐKXĐ:\(x\ge2\)

\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)

b,ĐKXĐ:\(x\in R\)

\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

c, ĐKXĐ:\(x\ge0\)

\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)

Nguyễn Việt Lâm, Nguyễn Lê Phước Thịnh giúp vs!

ĐKXĐ: \(x\ge2\).

Với \(x\ge2\) ta có \(VP\le2;VT\ge2\).

Do đó nghiệm của pt là \(x=2\).