Question

Cho biểu thức

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

a, Tìm ĐKXĐ và rút gọn

b, Tính A khi x=\(3-2\sqrt{2}\)

c, Tìm x để A=\(\frac{1}{4}\)

d, Tìm x nguyên để A nguyên

Answer 1

\(a;Đk:x\ne1\)

\(A=\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}-\frac{2}{x-1}=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2}{x-1}=\frac{\sqrt{x}-1}{x-1}+\frac{\sqrt{x}+1}{x-1}-\frac{2}{x-1}=\frac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2}{\sqrt{x}+1}\) \(b;x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}-1\Rightarrow A=\frac{2}{\sqrt{2}}=\sqrt{2}\)

\(c;A=\frac{1}{4}=\frac{2}{\sqrt{x}+1}\Leftrightarrow\sqrt{x}+1=8\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\)