a: \(x=\dfrac{4\left(\sqrt{3}-2\right)-4\left(\sqrt{3}+2\right)}{3-4}=-4\left(\sqrt{3}-2\right)+4\left(\sqrt{3}+2\right)\)
\(=-4\sqrt{3}+8+4\sqrt{3}+8=16\)
Khi x=16 thì \(A=\dfrac{2\cdot4+1}{4}=\dfrac{9}{4}\)
b: \(B=\dfrac{x-3\sqrt{x}+4-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
P=B:A
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{2\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}\)
c: |P|>P
=>P<0
=>\(\sqrt{x}-2< 0\)
=>0<=x<4
mà x>0 và x<>4
nên \(x\in\left\{1;2;3\right\}\)
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