Question

Cho biểu thức:

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

a) Tìm điều kiện của x để B có nghĩa

b) Thu gọn B

c) Tính x để B=3

Answer 1

ĐKXĐ: \(x\ge0;x\ne1\)

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

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

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

\(B=3\Rightarrow\sqrt{x}-1=3\Rightarrow\sqrt{x}=4\Rightarrow x=16\)