Giải hệ đẳng cấp: \(\hept{\begin{cases}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left(y+\sqrt{xy}+x-x^2\right)=4\end{cases}}\)
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\(\frac{144}{x+2}-\frac{100}{x}=2\)
\(\frac{144}{x-2}.x\left(x+2\right)-\frac{100}{x}.x\left(x+2\right)=2.x\left(x+2\right)\)
144x - 100(x + 2) = 2.x(x + 2)
x = 10
=> x = 10
K chắc nhá :w
\(ĐK:x\ne-2;x\ne0\)
\(\frac{144}{x+2}-\frac{100}{x}=2\)
\(\Leftrightarrow\frac{144}{x+2}-\frac{100}{x}-2=0\Leftrightarrow\frac{144x-100x-200-2x^2-4x}{\left(x+2\right)x}=0\)
\(\Leftrightarrow\frac{40x-2x^2-200}{\left(x+2\right)x}=0\Leftrightarrow40x-2x^2-200=0\Leftrightarrow20x-x^2-200=0\)
\(\Leftrightarrow-\left(20x-x^2-200\right)=0\Leftrightarrow x^2-20x+200=0\)
\(\Leftrightarrow\left(x-10\right)^2+100=0\left(\text{vô lí}\right)\)
\(\text{Vậy: pt vô nghiệm}\)
\(P=\frac{4\sqrt{x}+3}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(P=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}+1}=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}\inℤ\Leftrightarrow x+4\sqrt{x}+3⋮\sqrt{x}\)
Giải tiếp nhé sau đó thử chọn :V
\(p=\frac{4\sqrt{x}+3}{x+\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\frac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+3}{\sqrt{x}}=1+\frac{3}{\sqrt{x}}\)
Để \(x\in Z\Rightarrow P\in Z\)
\(\Rightarrow\sqrt{x}\inƯ\left(3\right)= \left\{-3;3\right\}\)
\(\Leftrightarrow x=9\left(t.mĐKXĐ\right)\)