ĐKXĐ : \(\hept{\begin{cases}2x-3\ge0\\x-1\ne0\\x-1\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x>1\end{cases}\Leftrightarrow}x\ge\frac{3}{2}.}\)

Vì x - 1 \(\ne\)0  => 2x - 3 = 0 

=> x = \(\frac{3}{2}\)( t /m )

Vây ........................

chắc vậy 

\(\sqrt{\frac{2x-3}{x-1}}=2\)

\(\frac{2x-3}{x-1}=4\)

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

\(2x-3-4x+4=0\)

\(-2x+1=0\)

\(x=-\frac{1}{2}\)

cái kia mk nhầm quên mất là có số 2 nha sorry 

a ) \(\sqrt{A^2}\)

ĐK : A\(\ge\)0

\(\sqrt{A^2}=A\)

b ) \(\sqrt{\frac{1}{A^2}}\)

ĐK : A\(\ge0\)

\(\sqrt{\frac{1}{A^2}}=\frac{\sqrt{1}}{A}\)

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

\(\Rightarrow x^4+4x^2+4=4\left(x^3+1\right)\)

\(\Rightarrow x^4+4x^2+4-4x^3-4=0\)

\(\Rightarrow x^4+4x^2-4x^3=0\)

\(\Rightarrow x^2\left(x^2-4x+4\right)=0\)

\(\Rightarrow\left(x-2\right)^2=0\)

\(\Rightarrow x=2\) 

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

\(\Leftrightarrow x^2+2=2\sqrt{x^3+1^3}\)

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

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

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

\(\Leftrightarrow x^4+4x^2+4=4x^3-4x^2+4x+4x^2-4x+4\)

\(\Leftrightarrow x^4+4x^2=4x^3\)

\(\Leftrightarrow x^4+4x^2-4x^3=0\)

\(\Leftrightarrow x^2\left(x^2+4-4x\right)=0\)

\(\Leftrightarrow x^2\left[x^2-2\left(x\right)\left(2\right)+2^2\right]=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy nghiệm phương trình là: {0; 2}

Tớ là nữ đây. Có đc hông thế?

1+1=3 khi làm sai.

Có đúng ko?