a: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)

\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+\sqrt{x}+22}{x-4}\)

d: Ta có: \(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-12}{x-9}\)

\(=\dfrac{\sqrt{x}-3+x+3\sqrt{x}+2\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x+6\sqrt{x}-15}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

Bạn đăng từng câu 1 nhé

giúp mik câu 1 với 2 nhé

\(A=\dfrac{\left(2-\sqrt[3]{x}\right)\left(4+2\sqrt[3]{x}+\sqrt[3]{x^2}\right)}{2+\sqrt[3]{x}}:\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}+\dfrac{\sqrt[3]{x^2}-2\sqrt[3]{x}+2\sqrt[3]{x}}{\sqrt[3]{x}-2}.\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x}+2\right)}{\sqrt[3]{x}\left(\sqrt[3]{x}+2\right)}\)

\(=\dfrac{\left(2-\sqrt[3]{x}\right)\left(4+2\sqrt[3]{x}+\sqrt[3]{x^2}\right)}{2+\sqrt[3]{x}}.\dfrac{2+\sqrt[3]{x}}{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}+\dfrac{\sqrt[3]{x}.\sqrt[3]{x}}{\sqrt[3]{x}-2}.\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x}+2\right)}{\sqrt[3]{x}\left(\sqrt[3]{x}+2\right)}\)

\(=2-\sqrt[3]{x}+\sqrt[3]{x}=2\)

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

\(\Leftrightarrow\dfrac{3+2x^2-2x-x+3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

\(\dfrac{1}{\sqrt{x}+2}>\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{5}>0\)

\(\Leftrightarrow\dfrac{5}{5\sqrt{x}+10}-\dfrac{\sqrt{x}+2}{5\sqrt{x}+10}>0\)

\(\Leftrightarrow\dfrac{5-\sqrt{x}-2}{5\sqrt{x}+10}>0\)

\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-3\right)}{5\sqrt{x}+10}>0\)

Mà: \(5\sqrt{x}+10\ge10>0\forall x\)

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

\(\Leftrightarrow x>9\)

_________

\(\dfrac{2}{\sqrt{x}+3}< \dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{4}{2\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{2\sqrt{x}+6}< 0\)

\(\Leftrightarrow\dfrac{4-\sqrt{x}-3}{2\sqrt{x}+6}< 0\)

\(\Leftrightarrow\dfrac{-\left(\sqrt{x}-1\right)}{2\sqrt{x}+6}< 0\)

Mà: \(2\sqrt{x}+6\ge6>0\forall x\)

\(\Leftrightarrow\sqrt{x}-1< 0\)

\(\Leftrightarrow\sqrt{x}< 1\)

\(\Leftrightarrow x< 1\)

\(\Leftrightarrow0\le x\le1\)

1: \(\Leftrightarrow\dfrac{3x-1}{x+2}=4\)

=>4x+8=3x-1

=>x=-9

2: \(\Leftrightarrow\dfrac{5x-7}{2x-1}=4\)

=>8x-4=5x-7

=>3x=-3

=>x=-1

3: ĐKXD: x>=0

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

=>\(x+\sqrt{x}-6=x-1\)

=>căn x=-1+6=5

=>x=25

4: ĐKXĐ: x>=0

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

=>x-2*căn x-3=x-4

=>-2căn x-3=-4

=>2căn x+3=4

=>2căn x=1

=>căn x=1/2

=>x=1/4

a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)

e cảm ơn ạ :33

Sửa đề: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{3}{2-\sqrt{x}}+\dfrac{3\sqrt{x}-2}{x-4}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{2\sqrt{x}-x}\right)\)

ĐKXĐ: x>0; x<>4

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

\(=\dfrac{x-2\sqrt{x}+3\sqrt{x}+6+3\sqrt{x}-2}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x+\sqrt{x}}\)

\(=\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{x+4\sqrt{x}+4}{x+3\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

\(\dfrac{8-x}{2+\sqrt[3]{x}}:\left(2+\dfrac{\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\left(\sqrt[3]{x}+\dfrac{2\sqrt[3]{x}}{\sqrt[3]{x}-2}\right).\left(\dfrac{\sqrt[3]{x^2}-4}{\sqrt[3]{x^2}+2\sqrt[3]{x}}\right)\)

\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}:\left(\dfrac{4+2\sqrt[3]{x}+\sqrt[3]{x^2}}{2+\sqrt[3]{x}}\right)+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\left(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x}+2\right)}{\sqrt[3]{x}\left(\sqrt[3]{x}+2\right)}\right)\)

\(\dfrac{\left(\sqrt[3]{x}-2\right)\left(\sqrt[3]{x^2}+2\sqrt[3]{x}+4\right)}{2+\sqrt[3]{x}}.\dfrac{2+\sqrt[3]{x}}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\dfrac{\sqrt[3]{x^2}}{\sqrt[3]{x}-2}.\dfrac{\sqrt[3]{x}-2}{\sqrt[3]{x}}\)

\(=\sqrt[3]{x}-2+\sqrt[3]{x}=2\sqrt[3]{x}-2\)