Question

Mn giúp em đề này vs ạ 

Answer 1

1: Thay x=16 vào A, ta được:

\(A=\frac{2}{\sqrt{16}+2}=\frac{2}{4+2}=\frac26=\frac13\)

2: \(B=\frac{1}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}\)

\(=\frac{\sqrt{x}+2+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

P=B:A

\(=\frac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{2}{\sqrt{x}+2}=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)

3: P<1

=>P-1<0

=>\(\frac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}<0\)

=>\(\frac{3}{\sqrt{x}-2}<0\)

=>\(\sqrt{x}-2<0\)

=>\(\sqrt{x}<2\)

=>0<=x<4

mà x là số nguyên

nên x∈{0;1;2;3}