Question

[Thái Quỳnh Hương]

Giải dùm ik cần gấp ạ

Answer 1

[Nguyễn Duy Long]

đừng dùng chat gbt nx

làm ơn

😤😤😤

Answer 2

[Nguyễn Lê Phước Thịnh]

ĐKXĐ: x>0; x<>1

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

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

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

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

b: Để A<0 thì \(\frac{x+\sqrt{x}}{\sqrt{x}-1}<0\)

=>\(\sqrt{x}-1<0\)

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

=>0<x<1

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

=>\(x-\sqrt{x}+2\sqrt{x}-2+2\) ⋮ \(\sqrt{x}-1\)

=>2⋮\(\sqrt{x}-1\)

=>\(\sqrt{x}-1\in\left\lbrace1;-1;2;-2\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2;3;-1\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2;3\right\rbrace\)

=>x∈{0;4;9}

Kết hợp ĐKXĐ, ta được: x∈{4;9}