Question

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

Rút gọn P và chúng minh \(P>\frac{2}{3}\)

Answer 1

P = 1 + \(\left(\frac{2x+\sqrt{x}-1}{1-x}-\frac{2x\sqrt{x}-\sqrt{x}+x}{1-x\sqrt{x}}\right)\).\(\left(\frac{x-\sqrt{x}}{2\sqrt{x}-1}\right)\) (x \(\ge\) 0; x \(\ne\) 1; x \(\ne\) \(\frac{1}{4}\))

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

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

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

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

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

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

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

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

Vậy ...

Ta có: 1 + x \(>\) 1 (x \(\ge\) 0; x \(\ne\) 1)

1 + \(\sqrt{x}\) + x > 1 (x \(\ge\) 0; x \(\ne\) 1)

\(\Rightarrow\) P = \(\frac{1+x}{1+\sqrt{x}+x}>1>\frac{2}{3}\) (đpcm)

Chúc bn học tốt!

Answer 2

Sửa lại chút phần chứng minh:

P > \(\frac{2}{3}\) \(\Leftrightarrow\) \(\frac{1+x}{1+\sqrt{x}+x}\) > \(\frac{2}{3}\) \(\Leftrightarrow\) \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}>\frac{-1}{3}\) \(\Leftrightarrow\) \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}+\frac{1}{3}>0\)

\(\Leftrightarrow\) \(\frac{-3\sqrt{x}+1+\sqrt{x}+x}{3\left(1+\sqrt{x}+x\right)}>0\) \(\Leftrightarrow\) \(\frac{1-2\sqrt{x}+x}{3\left(1+\sqrt{x}+x\right)}>0\) \(\Leftrightarrow\) \(\frac{\left(1-\sqrt{x}\right)^2}{3\left(1+\sqrt{x}+x\right)}>0\)

Ta thấy: x \(\ge\) 0 \(\Leftrightarrow\) \(\sqrt{x}\) \(\ge\) 0 \(\Leftrightarrow\) 1 + \(\sqrt{x}\) + x > 0 \(\Leftrightarrow\) 3(1 + \(\sqrt{x}\) + x) > 0 (1)

(1 - \(\sqrt{x}\))² \(\ge\) 0 với mọi x \(\ge\) 0

Dấu "=" xảy ra \(\Leftrightarrow\) x = 1 (KTM vì x \(\ne\) 1)

Vậy (1 - \(\sqrt{x}\))² \(\ge\) 0 với mọi x \(\ge\) 0; x \(\ne\) 1 (2)

Từ (1) và (2) (kết hợp với đk) \(\Rightarrow\) \(\frac{\left(1-\sqrt{x}\right)^2}{3\left(1+\sqrt{x}+x\right)}\) > 0 với mọi x \(\ge\) 0; x \(\ne\) 1; x \(\ne\) \(\frac{1}{4}\)

hay P > \(\frac{2}{3}\) với mọi x \(\ge\) 0; x \(\ne\) 1; x \(\ne\) \(\frac{1}{4}\) (đpcm)

Chúc bn học tốt! (phần rút gọn đúng r nha, mk chỉ sửa lại phần cm thôi!)