Question

cho N=1:\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+1+\sqrt{x}}-\dfrac{\sqrt{x}+1}{x-1}\right)\)

a. rút gọn N

b.So sánh N với 3

Answer 1

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

\(N=1:\left(\dfrac{x+2}{\sqrt{x^3}-1}+\dfrac{\sqrt{x}+1}{x+1+\sqrt{x}}-\dfrac{\sqrt{x}+1}{x-1}\right)\)

Điều kiện:x\(\ge\)0,x\(\ne\)1

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

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

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

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

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

\(=1:\left(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\)

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

Answer 2

a: \(N=1:\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

b: \(N-3=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)

=>N>3