Question

a) Tìm ĐKXĐ và rút gọn A:

A = \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\) - \(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}\)

b) Đặt B = A + x - 1. Tìm GTNN của B

Answer 1

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

Answer 2

\(B=A+x-1\)

\(=x-2\sqrt{x}-1\)

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

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

Vậy GTNN của B là -2 khi \(x=1\)