\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-5.\left(\dfrac{1}{2}\right)^3+3\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5-5\left(\dfrac{1}{2}\right)^3+6\left(\dfrac{1}{2}\right)^2+\dfrac{2}{2}+5\)

\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5.1}{8}+\dfrac{3.1}{4}+6-\dfrac{5.1}{8}+\dfrac{6.1}{4}+6\)

\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=-\dfrac{5}{8}+\dfrac{3}{4}+6-\dfrac{5}{8}+\dfrac{3}{2}+6\)

\(P\left(\dfrac{1}{2}\right)+Q\left(\dfrac{1}{2}\right)=13\)

\(Q\left(x\right)-P\left(x\right)=6\)

\(-5x^3+6x^2+2x+5+5x^3-3x^2-2x-5=6\)

\(3x^2=6\)

\(x^2=2\)

\(=>x=\pm\sqrt{2}\)

\(x^3+3x^2+x+a=x^2\left(x-2\right)+5x\left(x-2\right)+11\left(x-2\right)+22+a=\left(x-2\right)\left(x^2+5x+11\right)+22+a⋮\left(x-2\right)\)

\(\Rightarrow22+a=0\Rightarrow a=-22\)

bn ghi bắng công thức đi

a, \(Chof\left(x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

- Lập bảng xét dấu :

Vậy \(\left\{{}\begin{matrix}f\left(x\right)>0\Leftrightarrow x\in\left(3;4\right)\\f\left(x\right)< 0\Leftrightarrow x\in\left(-\infty;3\right)\cup\left(4;+\infty\right)\\f\left(x\right)=0\Leftrightarrow x\in\left\{3;4\right\}\end{matrix}\right.\)

b, \(f\left(x\right)=\left(x-1\right)\left(x+6\right)\)

( Làm tương tự câu a )

\(3x^2-6x-5x+5x^2-8x^2+24\)

\(=\left(3x^2+5x^2-8x^2\right)-\left(6x+5x\right)+24\)

\(=-11x+24\)

\(f\left(x\right)=3\Leftrightarrow\left|x-1\right|+2=3\Leftrightarrow\left|x-1\right|=1\\ \Leftrightarrow\left[{}\begin{matrix}x-1=1\\1-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

\(\)áp dụng BĐT AM-GM(BÀi này ko có Max chỉ có Min)

\(=>\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{xy}}=\dfrac{2}{\sqrt{xy}}\)

\(=>\dfrac{1}{2}\ge\dfrac{2}{\sqrt{xy}}=>\sqrt{xy}\ge4\)

\(=>S=\sqrt{x}+\sqrt{y}\ge2\sqrt{4}=4\)

dấu"=" xảy ra<=>x=y=4