1,Ta có: \(A=a^3+b^3+ab\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+ab\)
\(=a^2-ab+b^2+ab\)
\(=a^2+b^2\)
\(=\left(a+b\right)^2-2ab\)
\(=1-2ab\)
Vì \(a+b=1\Rightarrow a=1-b\)
Khi đó \(A=1-2\left(1-b\right)b\)
\(=1-2b-2b^2\)
\(=2\left(b^2-b+\dfrac{1}{4}\right)+\dfrac{1}{2}\)
\(=2\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)
Vì \(2\left(b-\dfrac{1}{2}\right)^2\ge0\Rightarrow A=2\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(\left(b-\dfrac{1}{2}\right)^2=0\Leftrightarrow b=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{2}\)
Vậy \(MinA=\dfrac{1}{2}\Leftrightarrow a=b=\dfrac{1}{2}\)
2, \(B=\dfrac{2}{6x-5-9x^2}=\dfrac{-2}{9x^2-6x+5}=\dfrac{-2}{\left(3x-1\right)^2+4}\)
Vì \(\left(3x-1\right)^2\ge0\Rightarrow\left(3x-1\right)^2+4\ge4\)
\(\Rightarrow\dfrac{1}{\left(3x-1\right)^2+4}\le\dfrac{1}{4}\)
\(\Rightarrow B=\dfrac{-2}{\left(3x-1\right)^2+4}\ge\dfrac{-2}{4}=\dfrac{-1}{2}\)
Dấu "=" xảy ra khi \(3x-1=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(MinB=\dfrac{-1}{2}\Leftrightarrow x=\dfrac{1}{3}\)
Cách khác :
Bài 1. Ta có : \(a^3+b^3+ab=\left(a+b\right)\left(a^2-ab+b^2\right)+ab=a^2+b^2\)
Áp dụng BĐT Bunhiacopxki , ta có :
\(\left(a^2+b^2\right)\left(1^2+1^2\right)\) ≥ \(\left(a+b\right)^2\)
⇔ \(a^2+b^2\) ≥ \(\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)
⇔ GTNN của \(a^2+b^2\) là \(\dfrac{1}{2}\) . Đẳng thức xảy ra khi : \(x=y=\dfrac{1}{2}\)
Bài 2. \(B=\dfrac{2}{6x-5-9x^2}=\dfrac{-2}{9x^2-6x+5}\)
\(B=\dfrac{-4}{2\left(9x^2-6x+5\right)}=\dfrac{-9x^2+6x-5+9x^2-6x+1}{2\left(9x^2-6x+5\right)}\)
\(B=\dfrac{-1}{2}+\dfrac{\left(3x-1\right)^2}{2\left(3x-1\right)^2+8}\)
Do : \(\dfrac{\left(3x-1\right)^2}{2\left(3x-1\right)^2+8}\) ≥ 0 ∀x
⇒ \(\dfrac{-1}{2}+\dfrac{\left(3x-1\right)^2}{2\left(3x-1\right)^2+8}\) ≥ \(\dfrac{-1}{2}\)
⇒ \(B_{Min}=\dfrac{-1}{2}\) ⇔ \(x=\dfrac{1}{3}\)
