Question

[Tung Nguyễn]

bài 1

Cho A=\(\frac{2x+1}{\sqrt{x}^3-1}-\frac{1}{\sqrt{x}-1}\)

B=\(\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\)

a)Tính B khi x=81

b)Rút gọn A

c)Tìm x thuộc Z để P thuộc Z với P=A:B

Answer 1

[Trần Việt Linh]

a) \(B=\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\left(ĐK:x\ge0\right)\)

\(=\frac{\sqrt{81}-3}{81+\sqrt{81}+1}=\frac{9-3}{81+9+1}=\frac{6}{91}\)

b) \(A=\frac{2x+1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\left(ĐK:x\ge0;x\ne1\right)\)

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

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

c) \(P=\frac{A}{B}\)

\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}:\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\left(ĐK:x\ge0;x\ne9\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}-3}\)

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

Vậy để P nguyên thì: \(\sqrt{x}-3\inƯ\left(3\right)\)

\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;-3;3\right\}\)

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

+) \(\sqrt{x}-3=1\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(tm\right)\)

+) \(\sqrt{x}-3=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

+) \(\sqrt{x}-3=3\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\left(tm\right)\)

Vậy...........