Question

[Ran]

  Mn làm giúp t câu 3 với

Answer 1

[Nguyễn Lê Phước Thịnh]

1: Thay x=36 vào A, ta được:

\(A=\dfrac{36-5}{\sqrt{36}}=\dfrac{31}{6}\)

2: \(B=\dfrac{2x+\sqrt{x}}{x-1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

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

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

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

3: \(P=A\cdot B=\dfrac{3x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-5}{\sqrt{x}}\)

\(=\dfrac{\left(x-5\right)\left(3\sqrt{x}+2\right)}{x-1}=\dfrac{\left(x-1\right)\left(3\sqrt{x}+2\right)-4\left(3\sqrt{x}+2\right)}{x-1}\)

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

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

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

=>\(9x-4⋮x-1\)

=>\(9x-9+5⋮x-1\)

=>\(5⋮x-1\)

=>\(x-1\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{2;0;6;-4\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;6\right\}\)

Khi x=2 thì \(P=3\sqrt{2}+2-\dfrac{4\left(3\sqrt{2}+2\right)}{2-1}\)

\(=3\sqrt{2}+2-4\left(3\sqrt{2}+2\right)=-3\left(3\sqrt{2}+2\right)\notin Z\)

=>Loại

Khi x=6 thì \(P=3\sqrt{6}+2-\dfrac{4\left(3\sqrt{6}+2\right)}{6-1}=3\sqrt{6}+2-\dfrac{4}{5}\left(3\sqrt{6}+2\right)\)

\(=\dfrac{1}{5}\left(3\sqrt{6}+2\right)\notin Z\)

=>Loại

Vậy: \(x\in\varnothing\)