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

= 1 nha a

939393:3=313131 nhoa bẹn

a) \(A=\left(x-y\right).\left(x^2+x+y\right)-x.\left(2x^2+2y^3\right)\)

\(=x^3+x^2+xy-x^2y-xy-y^2-2x^3-2xy^3\)

\(=-x^3-y^2-2xy^3\)

b) Ta thay \(x=-1;y=-5\)

\(-x^3-y^2-2xy^3\)

\(=-\left(-1\right)^3-\left(-5\right)^2-2.\left(-1\right).\left(-5\right)^3\)

\(=1-25-250\)

\(=-274\)

Ta có: \(a+b+c=0\)

\(\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)

\(\Rightarrow a^3+b^3+3ab.\left(a+b\right)=-c^3\)

\(\Rightarrow a^3+b^3+3ab.\left(-c\right)=-c^3\)

\(\Rightarrow a^3+b^3-3abc=-c^3\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

Vậy ta có điều cần phải chứng minh.

\(A=-x^2-10x+25\)

\(=-x^2-10x-25+50\)

\(=-\left(x+5\right)^2+50\le50\)

Đẳng thức khi x = -5

b) \(-x^2-8x+1\)

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

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

Đẳng thức khi x = -4

\(xy-y^2-3x+3y\)

\(=y\left(x-y\right)-3\left(x-y\right)\)

\(=\left(y-3\right)\left(x-y\right)\)