Question

[Lã thị huyền chinh]

Giúp e với ạ mn ơi

Answer 1

[Edogawa Conan]

ĐK: x > 0; x \(\ne\)1

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

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

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

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

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

b) Với x > 0 và x khác 1

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

Để A nhận giá trị nguyên <=> \(\frac{2}{\sqrt{x}-1}\)nguyên <=> \(2⋮\left(\sqrt{x}-1\right)\)

<=> \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Lập bảng: 

\(\sqrt{x}-1\)	1	-1	2	-2
x	4(tm)	0 (tm)	9 (tm)	không có giá trị tm

Vậy ...