Question

Cho biểu thức B=(\(\frac{2x+1}{\sqrt{x^3-1}}-\frac{\sqrt{x}}{x+\sqrt{x}+1}\))(\(\frac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\)) Với x ≥ 0 và x ≠ 0

a/ Rút gọn B

b/ Tìm x để B=3

Answer 1

Sửa đề nha: \(\sqrt{x^3-1}\) thành \(\sqrt{x^3}-1\)

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

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

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

b/ Để B= 3\(\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=16\)