Question

1. Tìm GTLN của biểu thức A= -3x+2\(\sqrt{x}\)+6 với x \(\ge\) 0

2. Tìm x, biết: \(\sqrt[3]{8}-\sqrt[3]{x}=-2\)

Answer 1

1.

\(A=-3x+2\sqrt{x}+6\\ =-3\left(x-\frac{2}{3}\sqrt{x}-2\right)\\ =-3\left[\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2-\frac{19}{9}\right]\\ =-3\left[\left(\sqrt{x}-\frac{1}{3}\right)^2\right]+\frac{19}{3}\le\frac{19}{3}\forall x\ge0\)

Vậy Max A = \(\frac{19}{3}\Leftrightarrow x=\frac{1}{9}\)

Answer 2

2.

\(\sqrt[3]{8}-\sqrt[3]{x}=-2\Leftrightarrow2-\sqrt[3]{x}=-2\\ \Leftrightarrow\sqrt[3]{x}=-4\\ \Leftrightarrow\left(\sqrt[3]{x}\right)^3=\left(-4\right)^3=-64\\ \Leftrightarrow x=-64\left(tm\right)\)