Question

giúp e vs ạ e cần gấp câu 1,2,3 ạ

Answer 1

1: Thay x=25 vào N, ta được:

\(N=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{5+1}{5}=\frac65\)

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

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

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

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

S=MN

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

3: S<-1

=>S+1<0

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

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

=>x-1<0

=>x<1

Kết hợp ĐKXĐ, ta được: 0<x<1