Xem có ai giúp hong nè
Bài 1
Rút gọn
\(P=\left(1+\dfrac{4}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
với \(x>0;x\ne4;x\ne1\)
Bài 2
Hệ phương trình
\(\left\{{}\begin{matrix}x+y=m-1\\x-y=m+3\end{matrix}\right.\) có nghiệm \(\left(x_0;y_0\right)\) thỏa mãn \(x_0=y_0^2\)
tìm giá trị của m
Bài 1:
\(P=\left(\dfrac{x-\sqrt{x}-2+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
\(\left\{{}\begin{matrix}x+y=m-1\\x-y=m+3\end{matrix}\right.\)
\(\Rightarrow x+y+x-y=m-1+m+3\)
\(\Rightarrow2x=2m+2\Rightarrow x=m+1\)
\(\Rightarrow x_0=m+1\) (1)
\(\left\{{}\begin{matrix}x+y=m-1\\x-y=m+3\end{matrix}\right.\)
\(\Rightarrow x+y-\left(x-y\right)=m-1-\left(m+3\right)\)
\(\Rightarrow2y=-4\Rightarrow y=-2\Rightarrow y_0=-2\Rightarrow y_0^2=4\) (2)
-Từ (1) và (2) suy ra:
\(m+1=4\Rightarrow m=3\)
