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đk: \(x\ge0\)và \(x\ne1\)
\(\Leftrightarrow P=\frac{x-1+\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-1}\right)}-\frac{2x-10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow P=\frac{x-1+x+\sqrt{x}-6-2x+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow P=\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)
để P > 0
\(\Leftrightarrow1>\sqrt{x}-1\)
\(\Leftrightarrow-\sqrt{x}>-2\)
\(\Leftrightarrow\sqrt{x}< 2\)
\(\Leftrightarrow x< 4\)
có sai xót mong m.n bỏ qa cho ♥
\(P=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\right)\)
ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
a, \(P=\left(\frac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)-\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}\right):\left(\frac{2\left(\sqrt{x}-1\right)^2}{x-1}\right)\)
\(\Leftrightarrow P=\left(\frac{x^2\sqrt{x}+x^2-x-\sqrt{x}-x^2\sqrt{x}+x^2-x+\sqrt{x}}{x\left(x-1\right)}\right):\left(\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow P=\frac{2x\left(x-1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)^2x\left(x-1\right)}\)
\(\Leftrightarrow P=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b,\(P=\frac{\sqrt{x}-1+2}{\sqrt{x}-1}\)
Để P thuộc Z
\(\Rightarrow2⋮\sqrt{x}-1\)
\(\Rightarrow\sqrt{x}-1\in\left(1;-1;2;-2\right)\)
\(\Leftrightarrow\sqrt{x}\in\left(2;0;3;-1\right)\)
\(\Leftrightarrow x=0\)(ko t/m đkxđ)
Vậy ko có x nguyên để P nguyên
\(B=\frac{2+\sqrt{x}}{x-4\sqrt{x}+4}:\left(\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}+\frac{6-x}{x+2\sqrt{x}}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{\sqrt{x}+2}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}+\frac{6-x}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)+\left(6-x\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{x\sqrt{x}-8+x+2\sqrt{x}+6\sqrt{x}-12-x\sqrt{x}+2x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{2+\sqrt{x}}{\left(\sqrt{x}-2\right)^2}:\left(\frac{3x+8\sqrt{x}-20}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(B=\frac{\sqrt{x}\left(2+\sqrt{x}\right)^2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2\left(3x+8\sqrt{x}-20\right)}\)
\(B=\frac{\sqrt{x}\left(2+\sqrt{x}\right)^2}{\left(\sqrt{x}-2\right)\left(3x+8\sqrt{x}-20\right)}\)
tới đây mình bí rồi cậu làm giúp mình đi
mại dzo