a: Khi x=9 thì \(A=\dfrac{2\cdot3-5}{3-2}=\dfrac{6-5}{1}=1\)
b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}}{4-x}-\dfrac{2}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-4\sqrt{x}-2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}-4\sqrt{x}-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\cdot\dfrac{2\sqrt{x}-5}{\sqrt{x}-2}=\dfrac{2\sqrt{x}-5}{\sqrt{x}+2}\)
P=1/2
=>\(\dfrac{2\sqrt{x}-5}{\sqrt{x}+2}=\dfrac{1}{2}\)
=>\(4\sqrt{x}-10=\sqrt{x}+2\)
=>\(3\sqrt{x}=12\)
=>căn x=4
=>x=16(nhận)
d: \(P-3=\dfrac{2\sqrt{x}-5}{\sqrt{x}+2}-3\)
\(=\dfrac{2\sqrt{x}-5-3\sqrt{x}-6}{\sqrt{x}+2}=\dfrac{-\sqrt{x}-11}{\sqrt{x}+2}< 0\)
=>P<3
e: P>-1
=>P+1>0
=>\(\dfrac{2\sqrt{x}-5+\sqrt{x}+2}{\sqrt{x}+2}>0\)
=>\(3\sqrt{x}-3>0\)
=>\(\sqrt{x}>1\)
=>x>1
Vậy: Số tự nhiên nhỏ nhất để P>-1 là x=2
e: P<0
=>\(2\sqrt{x}-5< 0\)
=>\(\sqrt{x}< \dfrac{5}{2}\)
=>0<=x<25/4
mà x là số nguyên tố
nên \(x\in\left\{2;3;5\right\}\)
