Bài 5:
a) \(a^5+b^5\ge a^3b^2+a^2b^3\)
\(\Leftrightarrow a^3\left(a^2-b^2\right)-b^3\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a^3-b^3\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\left(a+b\right)\ge0\) (luôn đúng với mọi a,b: \(a+b\ge0\))
Dấu "=" xảy ra \(\left\{{}\begin{matrix}a=b\\a+b\ge0\end{matrix}\right.\)
b) \(\dfrac{4}{x}+\sqrt{x-\dfrac{1}{x}}=x+\sqrt{2x-\dfrac{5}{x}}\)
(đk: \(x\ne0;x-\dfrac{1}{x}\ge0;2x-\dfrac{5}{x}\ge0\))
Đặt \(a=\sqrt{x-\dfrac{1}{x}},b=\sqrt{2x-\dfrac{5}{x}}\left(a,b\ge0\right)\)
\(\Rightarrow a^2-b^2=\dfrac{4}{x}-x\)
Pttt:\(a^2-b^2+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)
\(\Leftrightarrow a-b=0\) (vì a+b+1>0 với mọi a,b không âm)
\(\Leftrightarrow a=b\)\(\Rightarrow\sqrt{x-\dfrac{1}{x}}=\sqrt{2x-\dfrac{5}{x}}\) \(\Leftrightarrow x-\dfrac{1}{x}=2x-\dfrac{5}{x}\)
\(\Leftrightarrow\dfrac{4}{x}-x=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
Vậy x=2
