Ta có -2x²+3x+1

=-2(x²-\(\frac{3}{2}x\))+1

=-2(x²-2.\(\frac{3}{4}x\)+\(\frac{9}{16}-\frac{9}{16}\))+1

=-2(x-\(\frac{3}{4}\))²+\(\frac{9}{8}\)+1

==-2(x-\(\frac{3}{4}\))²+\(\frac{17}{8}\)

Đếnđây chắc bn tự lm đc r.tk mk nha

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP

óc người ak

don"t help

Đặt A = -2x^2 + 3x + 1

=> 2A = -4x^2 + 6x + 2

2A = - (4x^2 -6x-2) = - [(2x)² - 2.1,5.2.x + (1,5)² - 4,25 ] = - [(2x-1,5)²  - 4,25] = - (2x-1,5)² + 4,25

Để 2A lớn nhất

mà \(-\left(2x-1,5\right)^2\le0\)

Dấu "=" xảy ra khi: 2x-1,5 = 0 => x = 0,75

=> giá trị lớn nhất của 2A = -(2.0,75 -1,5)² + 4,25 = 4,25

=> ----------------------- A  = 4,25 : 2 = 2,125

KL:...

đặt k là cah hay nhat bn ak

\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2-a^3-b^3-c^3+4abc\)

\(=a\left(b-c\right)^2-a^3+4abc+b\left(c-a\right)^2-b^3+c\left(a-b\right)^2-c^3\)

\(=a\left[\left(b-c\right)^2+4bc-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)

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

\(=a\left(b+c+a\right)\left(b+c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)\)

\(=\left(b+c-a\right)\left[a\left(b+c+a\right)+b\left(c-a-b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)

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

\(=\left(b+c-a\right)\left[c\left(a+b\right)+\left(a-b\right)\left(a+b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)

\(=\left(b+c-a\right)\left(a+b\right)\left(a-b+c\right)+c\left(a-b+c\right)\left(a-b-c\right)\)

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

\(=\left(a-b+c\right)\left(b+a-c\right)\left(b-a+c\right)\)