Question

[Lê Phương Mai]

làm đi các bạn, mại zoooooo:")

Answer 1

[ILoveMath]

Bài 5:

\(a,\dfrac{a^2+b^2}{2}\ge ab\\ \Leftrightarrow a^2+b^2\ge2ab\\ \Leftrightarrow a^2-2ab+b^2\ge0\\ \Leftrightarrow\left(a-b\right)^2\ge0\left(luôn.đúng\right)\)

Dấu "=" xảy ra\(\Leftrightarrow a=b\)

\(b,\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2\\ \Leftrightarrow\dfrac{a^4+b^4}{a^2b^2}\ge2\\ \Leftrightarrow a^4+b^4\ge2a^2b^2\\ \Leftrightarrow a^4-2a^2b^2+b^4\ge0\\ \Leftrightarrow\left(a^2-b^2\right)^2\ge0\)

Dấu "=" xảy ra\(\Leftrightarrow a=b\)

Bài 6:

\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\\ \Rightarrow\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\\ \Rightarrow\left(x^2+5x-6\right)\left(x^2+5x+6\right)\\ \Rightarrow A=\left(x^2+5x\right)^2-36\ge-36\)

Dấu "=" xảy ra \(\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy \(A_{min}=-36\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 5: 

a: Ta có: \(\left(a-b\right)^2>=0\)

\(\Leftrightarrow a^2+b^2-2ab>=0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow\dfrac{a^2+b^2}{2}>=ab\)(đpcm)

b: \(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2\sqrt{\dfrac{a^2}{b^2}\cdot\dfrac{b^2}{a^2}}=2\)