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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a) \(A=2x^2-10x+2021=2\left(x^2-5x+\frac{25}{4}\right)+\frac{4017}{2}\)

\(=2\left(x-\frac{5}{2}\right)^2+\frac{4017}{2}\ge\frac{4017}{2}\)

Dấu \(=\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

b) \(B=x^2-7x-2008=x^2-7x+\frac{49}{4}-\frac{8081}{4}=\left(x-\frac{7}{2}\right)^2-\frac{8081}{4}\ge-\frac{8081}{4}\)

Dấu \(=\)khi \(x-\frac{7}{2}=0\Leftrightarrow x=\frac{7}{2}\)

c) \(C=2021-10x-x^2=-\left(x^2+10x+25\right)+2046=-\left(x+5\right)^2+2046\le2046\)

Dấu \(=\)khi \(x+5=0\Leftrightarrow x=-5\).

d) \(D=2008+10x-2x^2=-2\left(x^2-5x+\frac{25}{4}\right)+\frac{4041}{2}=-2\left(x-\frac{5}{2}\right)^2+\frac{4041}{2}\le\frac{4041}{2}\)

Dấu \(=\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).