Rút gọn biểu thức. Tìm x để P = 2

c) P = x(x - 1) = x² - x

\(=\ x^2-x\ +\frac{1}{4}-\frac{1}{4}\)

\(=\ \left(x-\frac{1}{2}\right)^2-\frac{1}{4}\)

Ta có \(\left(x-\frac{1}{2}\right)^2\ge0\) mọi x

<=> \(\left(x-\frac{1}{2}\right)^2-\frac{1}{4} \ge\frac{-1}{4}\) mọi x

<=> P \(\ge-\frac{1}{4}\)

Dấu "=" xảy ra khi x = 1/2

Vậy Min P = -1/4 khi x = 1/2

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

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

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

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

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

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

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

b) Để P = 2

<=> x(x - 1) = 2

<=> x² - x - 2 = 0

<=> (x - 2) (x + 1) = 0

<=> x = 2 hoặc x = -1