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Đk: x \(\ge\)0; x \(\ne\)1; x \(\ne\)9
1) \(B=\left(\frac{2x+3}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)
\(B=\frac{2x+3-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\)
\(B=\frac{-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(B=\frac{-\left(x+2\sqrt{x}-\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+2}{3-\sqrt{x}}\)
2. \(B=\frac{\sqrt{x}+2}{3-\sqrt{x}}=\frac{-\left(3-\sqrt{x}\right)+5}{3-\sqrt{x}}=-1+\frac{5}{3-\sqrt{x}}\)
Để B \(\in\)Z <=> 5 \(⋮\)\(3-\sqrt{x}\)
<=> \(3-\sqrt{x}\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Do \(3-\sqrt{x}\le\)3 => 3 - \(\sqrt{x}\)\(\in\){1; -1; -5}
Lập bảng:
| \(3-\sqrt{x}\) | 1 | -1 | -5 |
| x | 4 | 16 | 64 |
Vậy ...
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)