Câu 1:
a: Thay x=9 vào A, ta được:
\(A=\dfrac{3-1}{3}=\dfrac{2}{3}\)
b: \(B=\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{1-5\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}+\dfrac{1-5\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}+1\right)+1-5\sqrt{x}-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2x+2\sqrt{x}+1-5\sqrt{x}-x+2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(2P^2=P\)
=>\(P\left(2P-1\right)=0\)
=>\(\left[{}\begin{matrix}P=0\\P=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=0\\\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\sqrt{x}-1=0\\2\left(\sqrt{x}-1\right)=\sqrt{x}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\2\sqrt{x}-2=\sqrt{x}+1\end{matrix}\right.\)
=>\(2\sqrt{x}-\sqrt{x}=2+1\)
=>\(\sqrt{x}=3\)
=>x=9(nhận)
