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:\(x\ge0\)

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

\(\Leftrightarrow x+1=3x+9\\ \Leftrightarrow2x=-8\\ \Leftrightarrow x=-4\)

ĐKXĐ: \(x\ge3\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{4}\left(loại\right)\end{matrix}\right.\)

a) ĐKXĐ: x <= 2

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

??

Đề kiểu gì vậy ?

b) ĐKXĐ: x >= -1

pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3

a) ĐKXĐ: x <= 2/3

Pt --> 2 - 3x = 4

<=> 3x = -2

<=> x = -2/3 (thỏa)

b) ĐKXĐ: x >= 2

Pt --> x^2 + 4x + 4 = x^2 - 4x + 4

<=> 8x = 0<=> x = 0(loại)

a: Ta có: \(\sqrt{2-3x}=2\)

\(\Leftrightarrow2-3x=4\)

\(\Leftrightarrow3x=-2\)

hay \(x=-\dfrac{2}{3}\)

b: Ta có: \(\sqrt{x^2+4x+4}=x-2\)

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

\(\Leftrightarrow x+2=2-x\left(x< -2\right)\)

\(\Leftrightarrow x=0\left(loại\right)\)

\(x^2+2x+4=3\sqrt{x^3+4x}\)đk \(x\ge0\)

\(x^2+2x+4=3\sqrt{x\left(x^2+4\right)}\)

đặt \(x^2+4=t\)

=> \(t+2x=3\sqrt{tx}\Leftrightarrow t^2-5tx+4x^2=0\)

\(\Leftrightarrow\left(t-x\right)\left(t-4x\right)=0\Leftrightarrow\orbr{\begin{cases}t=x\\t=4x\end{cases}}\)

nếu t=x phương trình trở thành \(x^2+4=x\Leftrightarrow x^2-x+4=0\Rightarrow ptvonghiem\)

nếu t=4x phương trinh trở thành \(x^2+4=4x\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

vậy x=2 là nghiệm của pt 

x=2

nhớ k cho nha