Câu 1:
1: Thay x=25 vào B, ta được:
\(B=\dfrac{1}{5-2}=\dfrac{1}{3}\)
2: P=A:B
\(=\left(\dfrac{x+2}{x-\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\right):\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{x+2-2\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{1}\)
\(=\dfrac{x+2-2x+4\sqrt{x}+x-1}{\sqrt{x}+1}=\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}\)
3:
Ta có: \(4\sqrt{x}+1>=1>0\forall x\) thỏa mãn ĐKXĐ
\(\sqrt{x}+1>=1>0\forall x\) thỏa mãn ĐKXĐ
=>\(P=\dfrac{4\sqrt{x}+1}{\sqrt{x}+1}>0\forall x\) thỏa mãn ĐKXĐ
\(P^2=P+2\)
=>\(P^2-P-2=0\)
=>(P-2)(P+1)=0
mà P+1>=1(do P>=0)
nên P-2=0
=>P=2
=>\(4\sqrt{x}+1=2\left(\sqrt{x}+1\right)=2\sqrt{x}+2\)
=>\(\sqrt{x}=\dfrac{1}{2}\)
=>x=1/4(nhận)
