\(f\left(x\right)=2x^3+ax^2+bx+c\)

Ta có: 

\(\hept{\begin{cases}f\left(-1\right)=-6\\f\left(1\right)=4\\f\left(2\right)=21\end{cases}}\Leftrightarrow\hept{\begin{cases}a-b+c=-4\\a+b+c=2\\4a+2b+c=5\end{cases}}\Leftrightarrow\hept{\begin{cases}a=0\\b=3\\c=-1\end{cases}}\)

Do đó \(f\left(x\right)=2x^3+3x-1\).

\(f\left(5\right)=264\).

\(B=-x^2+8x+1\)

\(=-x^2+8x-16+17\)

\(=-\left(x-4\right)^2+17\le17\)

helppp

\(x^3-19x-30=0\)

\(\Rightarrow x^3-25x+6x-30=0\)

\(\Rightarrow x\left(x^2-25\right)+6\left(x-5\right)=0\)

\(\Rightarrow x\left(x^2-5^2\right)+6\left(x-5\right)=0\)

\(\Rightarrow x\left(x-5\right)\left(x+5\right)+6\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)[x\left(x+5\right)+6]=0\)

\(\Rightarrow\left(x-5\right)\left(x^2+5x+6\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x^2+2x+3x+6\right)=0\)

\(\Rightarrow\left(x-5\right)[x\left(x+2\right)+3\left(x+2\right)]=0\)

\(\Rightarrow\left(x+2\right)\left(x+3\right)\left(x-5\right)=0\)

Trường hợp 1: \(x+2=0\Rightarrow x=-2\)

Trường hợp 2: \(x+3=0\Rightarrow x=-3\)

Trường hợp 3: \(x-5=0\Rightarrow x=5\)

Vậy giá trị lớn nhất của \(x=5\)

\(x^3-19x-30=0\)

\(\Leftrightarrow x^3-5x^2+5x^2-25x+6x-30=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+3x+2x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x+2\right)=0\)

\(\Leftrightarrow x-5=0\)hoặc \(x+3=0\)hoặc \(x+2=0\)

\(\Leftrightarrow x=5\)hoặc \(x=-3\)hoặc \(x=-2\).

Vậy \(x=5\)là giá trị thỏa mãn ycbt. 

\(x^2-8x+7\)

\(=x^2-x-7x+7\)

\(=x\left(x-1\right)-7\left(x-1\right)\)

\(=\left(x-7\right)\left(x-1\right)\)

1) A = x^2 + 3x + 10
=x^2 + 2x.3/2 + 9/4 + 31/4
=(x+3/2)^2 + 31/4
mà (x+ 3/2 )^2 >=0 với mọi x
=> Min (x+3/2)^2 = 0
=> Min (x+3/2)^2 + 31/4 = 31/4
Vậy GTNN của A là 31/4 khi:
x+ 3/2 = 0 <=> x= - 3/2

\(B=x^2-3x+10\)

\(=x^2-3x+\frac{9}{4}+\frac{31}{4}\)

\(=\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)

\(\left(x^2-1\right)\left(x^2+2x\right)\)

\(=\left(x^2+2x\right).x^2-1\left(x^2+2x\right)\)

\(=x^4+2x^3-1\left(x^2+2x\right)\)

\(=x^4+2x^3-\left(2x^3+2x\right)\)

\(=x^4+2x^3-x^2-2x\)