Đặt \(A=-5x^2+2x+1\)

\(A=\left(-5x^2+2x-\frac{1}{5}\right)+\frac{6}{5}\)

\(A=-5\left(x^2-\frac{2}{5}x+\frac{1}{25}\right)+\frac{6}{5}\)

\(A=-5\left(x-\frac{1}{5}\right)^2+\frac{6}{5}\le\frac{6}{5}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(-5\left(x-\frac{1}{5}\right)^2=0\)\(\Leftrightarrow\)\(x=\frac{1}{5}\)

Vậy GTLN của \(A\) là \(\frac{6}{5}\) khi \(x=\frac{1}{5}\)

Chúc bạn học tốt ~ 

Gọi biểu thức trên là T

Ta có: \(T=-5x^2+2x+1=-\left(5x^2-2x\right)+1\)

\(=-\left(5x^2-2x+1\right)+\frac{6}{5}=-\left(5x+1\right)^2+\frac{6}{5}\)

Vì \(-\left(5x+1\right)^2\le0\forall x\) nên \(T=-\left(5x+1\right)^2+\frac{6}{5}\le\frac{6}{5}\)

Dấu "=" xảy ra khi \(-\left(5x+1\right)^2=0\Leftrightarrow x=\frac{1}{5}\)

Vậy \(T_{max}=\frac{6}{5}\Leftrightarrow x=\frac{1}{5}\)

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

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

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

\(=\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

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

\(=a^3+3a^2b+3ab^2+b^3+c^3-3abc-3a^2b-3ab^2\)

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

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)

d) \(x^2-y^2-2x+2y\)

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

\(=\left(x-1\right)^2-\left(y-1\right)^2\)

\(=\left(x-1-y+1\right)\left(x-1+y-1\right)\)

\(=\left(x-y\right)\left(x+y-2\right)\)

\(4xy^2-12x^2y+8xy\)

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

\(3x^2-6xy+3y^2-12z^2\)

\(=3.\left(x^2-2xy+y^2-4z^2\right)\)

\(=3.\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=3.\left(x-y-2z\right)\left(x-y+2z\right)\)

\(x^4y^4+4=\left[\left(x^2y^2\right)^2+2..x^2y^2.2+2^2\right]-\left(2xy\right)^2\)

\(=\left(x^2y^2+2\right)^2-\left(2xy\right)^2\)

\(=\left(x^2y^2+2-2xy\right)\left(x^2y^2+2+2xy\right)\)

Vì tổng số trận đấu là 10 trận khi đó \(\frac{x(x-1)}{2}=10\)

Ta có : \(\frac{x(x-1)}{2}=10\)

           \(\Rightarrow x(x-1)=10\cdot2\)

           \(\Rightarrow x(x-1)=20\)

Do 20 = 4.5 nên có 5 đội tham gia thi đấu

\(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x+y\right)\left(x-y\right)\)

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

\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)

\(=2y^2-10xy\)