Question

1.Giải phương trình

a) \(\left(x^2-6x+8\right)\left(x^2-10x+8\right)-12x-39=0\)

b) \(x^2+5x=2\sqrt{x^2+5x-2}-2\)

2. cho a, b>0 và a+b=1. Tìm GTNN của \(A=\frac{1}{ab}+\frac{1}{a^2}+\frac{1}{b^2}\)

Answer 1

1.b)

ĐKXĐ: \(x^2+5x-2\ge0\)

PT \(\Leftrightarrow x^2+5x-2-2\sqrt{x^2+5x-2}+1=-3\)

\(\Leftrightarrow\left(\sqrt{x^2+5x-2}-1\right)^2=-3\)(vô nghiệm)

Answer 2

2.

\(A=\frac{1}{ab}+\frac{1}{a^2}+\frac{1}{b^2}\)\(=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{2ab}+\left(\frac{1}{a}+\frac{1}{b}\right)^2\)

Ta có: \(2ab\le\frac{\left(a+b\right)^2}{2}=\frac{1}{2}\)\(\Rightarrow\frac{1}{2ab}\ge2\)

\(\left(\frac{1}{a}+\frac{1}{b}\right)^2\ge\left(\frac{4}{a+b}\right)^2=16\)

\(\Rightarrow A\ge18\). Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)

Vậy min A=18\(\Leftrightarrow a=b=\frac{1}{2}\)