Question

Cho A=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{1-x}\) với x≥0,x≠1

a) Rút gọn A

b) Tìm m để phương trình mA=\(\sqrt{x}-2\) có 2 nghiệm phân biệt

c) Tìm x để A nhận giá trị nguyên

Answer 1

a: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{1-x}\)

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

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

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

c: Để A nguyên thì \(2\sqrt{x}+1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-1\) ⋮\(\sqrt{x}+1\)

=>-1⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1=1\)

=>x=0(nhận)