Question

[Vo Thi Minh Dao]

cho bieu thuc A=\(\left(\dfrac{6x+4}{3\sqrt{3x^3}-8}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\dfrac{1+3\sqrt{3x^3}}{1+\sqrt{3x}}-\sqrt{3x}\right)\)

a.rut gon bieu thuc A

b. tim cac gia tri nguyen x de A nguyen

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(A=\left(\dfrac{6x+4}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}-\dfrac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\dfrac{1+\left(\sqrt{3x}\right)^3}{1+\sqrt{3x}}-\sqrt{3x}\right)\)

\(=\dfrac{6x+4-3x+2\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\cdot\left(1-\sqrt{3x}\right)^2\)

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

b: Để A là số nguyên thì \(3x-2\sqrt{3x}+1⋮\sqrt{3x}-2\)

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

=>\(3x\in\left\{9;1;25\right\}\)

hay x=3