Bài 1:
1: Thay x=25 vào A, ta được:
\(A=\dfrac{5+2}{5}=\dfrac{7}{5}\)
2: \(B=\dfrac{1}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}}+\dfrac{3\sqrt{x}-8}{x-2\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2\left(\sqrt{x}-2\right)+3\sqrt{x}-8}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)
\(=\dfrac{4\sqrt{x}-8-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{2\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{2}{\sqrt{x}}\)
3: M=B:A
\(=\dfrac{2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{2}{\sqrt{x}+2}\)
\(M=\dfrac{2}{3}\)
=>\(\dfrac{2}{\sqrt{x}+2}=\dfrac{2}{3}\)
=>\(\sqrt{x}+2=3\)
=>\(\sqrt{x}=1\)
=>x=1(nhận)
Bài 2:
1: Thay x=16 vào P, ta được:
\(P=\dfrac{4-2}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
2: \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}+\dfrac{7\sqrt{x}-3}{x-9}-\dfrac{3\sqrt{x}-x}{\sqrt{x}-3}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)+7\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x-3\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{x-4\sqrt{x}+3+7\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x-3\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{x-3\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{x-3\sqrt{x}}{\sqrt{x}-3}=\dfrac{x-2\sqrt{x}}{\sqrt{x}-3}\)
3: M=PQ
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-3}\)
Để M>=0 thì \(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-3}>=0\)
=>\(\sqrt{x}-3>0\)
=>\(\sqrt{x}>3\)
=>x>9
