a) Đặt  \(A=-x^2+9x-12\)

\(-A=x^2-9x+12\)

\(-A=\left(x^2-9x+\frac{81}{4}\right)-\frac{33}{4}\)

\(-A=\left(x-\frac{9}{2}\right)^2-\frac{33}{4}\)

Mà  \(\left(x-\frac{9}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge-\frac{33}{4}\Leftrightarrow A\le\frac{33}{4}\)

Dấu "=" xảy ra khi :  \(x-\frac{9}{2}=0\Leftrightarrow x=\frac{9}{2}\)

Vậy  \(A_{Max}=\frac{33}{4}\Leftrightarrow x=\frac{9}{2}\)

b) Đặt \(B=2x^2+10x-1\)

\(B=2\left(x^2+5x+\frac{25}{4}\right)-\frac{29}{4}\)

\(B=2\left(x+\frac{5}{2}\right)^2-\frac{29}{4}\)

Mà  \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\Rightarrow2\left(x+\frac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow B\ge-\frac{29}{4}\)

Dấu "=" xảy ra khi :  \(x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy  \(B_{Min}=-\frac{29}{4}\Leftrightarrow x=-\frac{5}{2}\)

c) Đặt  \(C=\left(2x+6\right)\left(x-1\right)\)

\(C=2x^2-2x+6x-6\)

\(C=2x^2+4x-6\)

\(C=2\left(x^2+2x+1\right)-8\)

\(C=2\left(x+1\right)^2-8\)

Mà  \(\left(x+1\right)^2\ge0\forall x\Rightarrow2\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow C\ge-8\)

Dấu "=" xảy ra khi :  \(x+1=0\Leftrightarrow x=-1\)

Vậy  \(C_{Min}=-8\Leftrightarrow x=-1\)

d) Đặt  \(D=3x-2x^2\)

\(-2D=4x^2-6x\)

\(-2D=\left(4x^2-6x+\frac{9}{4}\right)-\frac{9}{4}\)

\(-2D=\left(2x-\frac{3}{2}\right)^2-\frac{9}{4}\)

Mà  \(\left(2x-\frac{3}{2}\right)^2\ge0\forall x\)

\(\Rightarrow-2D\ge-\frac{9}{4}\)

\(\Leftrightarrow D\le\frac{9}{8}\)

Dấu "=" xảy ra khi :  \(2x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{4}\)

Vậy  \(D_{Max}=\frac{9}{8}\Leftrightarrow x=\frac{3}{4}\)

\(1,3x\left(x-2\right)-2\left(x-2\right)\)

\(=\left(x-2\right)\left(3x-2\right)\)

\(2,5x\left(x-3\right)+\left(x-3\right)\)

\(=\left(5x-1\right)\left(x-3\right)\)

\(3,5x\left(x-3\right)-x+3\)

\(=\left(5x-1\right)\left(x-3\right)\)

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

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

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

1) 3x(x-2) - 2(x-2)=(x-2)(3x-2)

2) 5x(x-3) +x-3=5x(x-3)+(x-3)=(x-3)(5x+1)

3) 5x(x-3) - x+3=5x(x-3) -(x-3)=(x-3)(5x-1)

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

Hok tốt nha!!!=.=

2018^3 -1 = (2018-1)(2018^2 + 2018+1)

2018^2 + 2019 = 2018^2 + 2018+1 

Vậy 2018^3 -1 / 2018^2 +2019 = 2018 -1= 2017

Chúc bạn học tốt.

a) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)

\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)

\(\Leftrightarrow-14x=14\)

\(\Leftrightarrow x=-1\)

b) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(\Leftrightarrow17x=-34\Rightarrow x=-2\)

con đường dài 60km