Question

Giúp mình với ạ

Answer 1

a: \(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}=\dfrac{2x-3\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)

b: Khi x=4-2căn 3 thì \(A=\dfrac{2\left(\sqrt{3}-1\right)-2}{\sqrt{3}-1+1}=\dfrac{2\sqrt{3}-4}{\sqrt{3}}=2-\dfrac{4\sqrt{3}}{3}\)

c: \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)

KhiA=-1/2 thì \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{-1}{2}\)

=>\(4\sqrt{x}-2=-\sqrt{x}-1\)

=>5căn x=1

=>x=1/25

d: Để A>0 thì 2 căn x-1>0

=>x>1/4

Để A<0 thì 2 căn x-1<0

=>0<x<1/4

e: Để A nguyên thì \(2\sqrt{x}+2-3⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\{1;3\right\}\)

=>\(x\in\left\{0;4\right\}\)