\(ĐKXĐ:x\ne\pm1\)

\(A=\left(1+\frac{\sqrt{x}}{x+1}\right)\div\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)

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

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

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

\(\Leftrightarrow A=\frac{x+\sqrt{x}+1}{x+1}:\frac{\sqrt{x}-1}{x+1}\)

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

b. Tìm giá trị của x sao cho A > 3 

Giúp mình ý tiếp với 

Phương trình sau <=> \(\left(1+3x+2x^2\right)\left(1+3x\right)=\left(1+3y+2x^2\right)\left(1+3y\right)\)

<=> \(\left(1+3x\right)^2+2x^2\left(1+3x\right)-\left(1+3y\right)^2-2x^2\left(1+3y\right)=0\)

<=> \(\left[\left(1+3x\right)^2-\left(1+3y\right)^2\right]+\left[2x^2\left(1+3x\right)-2x^2\left(1+3y\right)\right]=0\)

<=> \(\left(3x-3y\right)\left(2+3x+3y\right)+2x^2\left(3x-3y\right)=0\)

<=> \(\left(3x-3y\right)\left(2+3x+3y+2x^2\right)=0\)

<=> \(\orbr{\begin{cases}x=y\\2x^2+3x+3y+2=0\end{cases}}\)

Với x = y ta có hệ : \(\hept{\begin{cases}x-5y=-20\\x=y\end{cases}}\Leftrightarrow x=y=5\)

Với \(2x^2+3x+3y+2=0\)ta có hệ: \(\hept{\begin{cases}x-5y=-20\\2x^2+3x+3y+2=0\end{cases}}\) hệ này đơn giản em tự giải tiếp!

He ..........me about it last night.

A.tells.                            B.have told.                             C.has told                       D.told

My house..........broken into last night

A.are.                           B.is.                                C.was.             D.were

He ..........me about it last night.

A.tells.                            B.have told.                             C.has told                       D.told

My house..........broken into last night

A.are.                           B.is.                                C.was.             D.were