bình phương vế trái lên nhé bạn ơi

Cm giúp mình với nhé. Mình cảm ơn ạ!

đề bài là j vậy bn

\(D=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

\(D=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3\sqrt{x}+1-\left(3\sqrt{x}-2\right)}\)

\(D=\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3\sqrt{x}+1-3\sqrt{x}+2}\)

\(D=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}\)

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

Sửa đề :

a) \(A=\left(\frac{x-\sqrt{x}}{x-\sqrt{x}-2}+\frac{4}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-\sqrt{x}-5}{x-\sqrt{x}-2}\right)\)

\(\Leftrightarrow A=\frac{x-\sqrt{x}+4\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{x-4-x+\sqrt{x}+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(\Leftrightarrow A=\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}\)

b) \(A=4\)

\(\Leftrightarrow\frac{x+3\sqrt{x}+4}{\sqrt{x}+1}=4\)

\(\Leftrightarrow x+3\sqrt{x}+4=4\sqrt{x}+4\)

\(\Leftrightarrow x-\sqrt{x}=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy \(A=4\Leftrightarrow x\in\left\{0;1\right\}\)

\(\frac{x}{x-1}-1=\frac{2x}{x+3}\) \(ĐKXĐ:x\ne1;x\ne-3\)

\(\Leftrightarrow\frac{x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow x^2+3x-\left(x^2+3x-x-3\right)=2x^2+6x\)

\(\Leftrightarrow x^2+3x-x^2-3x+x+3=2x^2+6x\)

\(\Leftrightarrow x^2-x^2-2x^2+3x-3x+x-6x+3=0\)

\(\Leftrightarrow-2x^2-5x+3=0\)

\(\Leftrightarrow-2x^2-6x+x+3=0\)

\(\Leftrightarrow-2x\left(x+3\right)+\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(1-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\1-2x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\left(ko.t.m\right)\\x=\frac{1}{2}\left(t.m\right)\end{cases}}\)

Vậy...

TÌM GIÁ TRỊ LỚN NHẤT, NHỎ NHẤT CỦA CÁC BIỂU THỨC SAU ( NẾU CÓ) :
A=X−−√+1X+1
B=3(X−−√−1)+73(X−1)+7
C=4X−2−−−−−√−34X−2−3
D=−2017x√+1−2017x+1
E=x+1√x√+2x+1x+2
F=x+2x−−√−5x+2x−5
G=1x2−4x+5√