Question

Cho bt:

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

a) Rút gọn

b) CMR bt P luôn luôn âm với mọi x

Answer 1

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

mà \(-\sqrt{x}< 0\) ( vì điều kiện xác định x > 0 ; x \(\ne\) -1 )

mặt khác \(x-\sqrt{x}+1=x-2\sqrt{x}\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

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

=> p \(< \) 0 => p luôn luôn âm với mọi x

Answer 2

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

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

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

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

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

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

Answer 3

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

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

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

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

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

b) Vì \(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{2\left(x\sqrt{x}+1\right)}>0\forall x\)

\(\Rightarrow\frac{-\sqrt{x}\left(\sqrt{x}+1\right)}{2\left(x\sqrt{x}+1\right)}< 0\)( đpcm )