Tìm a, b để: f(x) = x^3 + ax^2 +bx - 1 chia hết cho g(x)= x^2 + x - 2

Ta có: \(g\left(x\right)=0\)

\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Để f(x) chia hết cho g(x) thì mọi nghiệm của g(x) đều là nghiệm của f(x) nên:

\(\hept{\begin{cases}f\left(1\right)=0\\f\left(-2\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=0\\4a-2b=9\end{cases}}\Leftrightarrow\hept{\begin{cases}2a+2b=0\\4a-2b=9\end{cases}}\)

\(\Leftrightarrow6a=9\Rightarrow a=\frac{3}{2}\Rightarrow b=-\frac{3}{2}\)

Vậy \(\hept{\begin{cases}a=\frac{3}{2}\\b=-\frac{3}{2}\end{cases}}\)

\(\left(x^2-2x-1\right)\left(x-3\right)=x^3-3x^2-2x^2+6x-x+3=x^3-5x^2+5x+3\)

Ta có: \(3x^2+y^2-2xy+2y-10x+2029\)

\(=\left[\left(x^2-2xy+y^2\right)-\left(2x-2y\right)+1\right]+\left(2x^2-8x+8\right)+2020\)

\(=\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]+2\left(x^2-4x+4\right)+2020\)

\(=\left(x-y-1\right)^2+2\left(x-2\right)^2+2020\ge2020\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\2\left(x-2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

Vậy Min = 2020 khi x = 2 và y = 1

Ta có: \(a^4+a^2+1\)

\(=\left(a^4+2a^2+1\right)-a^2\)

\(=\left(a^2+1\right)^2-a^2\)

\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)

\(a^4+a^2+1=a^4+2a^2-a^2+1\)

\(=\left(a^4+2a^2+1\right)-a^2\)

\(=\left(a^2+1\right)^2-a^2\)

\(=\left(a^2+a+1\right)\left(a^2-a+1\right)\)

\(B=-2x^2+10x-8=-2x^2+10x-\frac{25}{2}+\frac{9}{2}\)

\(=-\left(2x^2-10x+\frac{25}{2}\right)+\frac{9}{2}\)

\(=-2\left(x^2-5x+\frac{25}{4}\right)+\frac{9}{2}\)

\(=-2\left[x^2-2.\frac{5}{2}x+\left(\frac{5}{2}\right)^2\right]+\frac{9}{2}\)

\(=-2\left(x-\frac{5}{2}\right)^2+\frac{9}{2}\)

Vì \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)\(\Rightarrow-2\left(x-\frac{5}{2}\right)^2\le0\forall x\)

\(\Rightarrow-2\left(x-\frac{5}{2}\right)^2+\frac{9}{2}\le\frac{9}{2}\forall x\)

hay \(B\le\frac{9}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\)\(\Leftrightarrow x=\frac{5}{2}\)

Vậy \(maxB=\frac{9}{2}\)\(\Leftrightarrow x=\frac{5}{2}\)

Ta có: 

\(B=-2x^2+10x-8\)

\(B=-2\left(x^2-5x+\frac{25}{4}\right)+\frac{9}{2}\)

\(B=-2\left(x-\frac{5}{2}\right)^2+\frac{9}{2}\le\frac{9}{2}\left(\forall x\right)\)

Dấu "=" xảy ra khi: x = 5/2

Vậy Max(B) = 9/2 khi x = 5/2