Question

Rút gọn:   \(\left(\frac{x\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{x+1}\right):\frac{\sqrt{x}-1}{x+1}\)

 

Answer 1

ĐKXĐ:

\(x\ge0\text{ và }\sqrt{x}-1\ge0\text{ và }x+1\ne0\)

\(\Leftrightarrow x\ge1\text{ và }x\ne-1\)

\(\left(\frac{x\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}+\frac{1}{x+1}\right):\frac{\sqrt{x}-1}{x+1}\)

\(=\left(\frac{x\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}+1}\)

\(=\left(\frac{x\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\right).\frac{x+1}{\sqrt{x}+1}\)

\(=\frac{x\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}.\frac{x+1}{\sqrt{x}+1}\)

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

Answer 2

ko bik làm hay làm biến đó