\(\left(3x+1\right)\left(2x-5\right)\)

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

\(=6x^2-15x+2x-5\)

\(=6x^2-13x-5\)

giúp tôi với ạ

A=1²-2²+3²-...+2019²

=1²+2²+...+2019²-2(2²+4²+...+2018²)

=B                       -8(1²+2²+...+1009²)

=B                       -8C

với B=1²+2²+...+2019²=1(2-1)+2(3-1)+...+2019(2020-1)

=1.2+2.3+...+2019.2020-(1+2+...+2019)

=D-2039190

với D=1.2+2.3+...+2019.2020

3D=1.2.3+2.3.(4-1)+...+2019.2020(2021-2018)

3D=1.2.3+2.3.4+...+2019.2020.2021-(1.2.3+2.3.4+...+2018.2019.2020)

3D=2019.2020.2021 suy ra D=2747468660

ta có :B=2747468660-2039190=2745429470

làm C như B ta có C=342923785

ta có: A=B-C=2745429470-8.342923785=2039190

vậy 1²-2²+3²-...+2019²=2039190

đúng thì cho mình nha

Ta có:\(1^2-2^2+3^2-4^2+5^2+...-2018^2+2019^2\)

\(=\left(2019^2-2018^2\right)+...+\left(3^2-2^2\right)+1^2\)

\(=\left(2019-2018\right)\left(2019-2018\right)+...+\left(3-2\right)\left(3+2\right)+1\)

\(=2019+2018+...+3+2+1\)

\(=\frac{2019\left(2019+1\right)}{2}\)

\(=2039190\)

\(C=x\left(x-3\right)\left(x+3\right)-\left(x+1\right)\left(x^2-7x\right)\)

\(=x\left(x^2-9\right)-x\left(x+1\right)\left(x-7\right)\)

\(=x^3-9x-x^3+7x^2-x^2+7x=6x^2-2x\)

\(=2x\left(3x-1\right)\)Thay x = 2/3 vào biểu thức trên ta được : 

\(2.\frac{2}{3}\left(3.\frac{2}{3}-1\right)=\frac{4}{3}\left(2-1\right)=\frac{4}{3}\)

Ta có:\(C=x\left(x-3\right)\left(x+3\right)-\left(x+1\right)\left(x^2-7x\right)\)

\(=x\left(x^2-9\right)-\left[x\left(x^2-7x\right)+1\left(x^2-7x\right)\right]\)

\(=x^3-9x-x^3+7x^2-x^2+7x\)

\(=-x^2-2x+7\)

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

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

Tại \(x=\frac{2}{3}\)thì giá trị của C là:

\(C=2.\frac{2}{3}\left(3.\frac{2}{3}-1\right)\)

\(=\frac{4}{3}\left(2-1\right)=\frac{4}{3}\)

\(C=x\left(x-3\right)\left(x+3\right)-\left(x+1\right)\left(x^2-7x\right)\)

\(=x\left(x^2-9\right)-\left(x^3-6x^2-7x\right)\)

\(=x^3-9x-x^2+6x^2+7x\)

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

\(=2x\left(3x+1\right)\)

với \(x=\frac{2}{3}\)phương trình C trở thành :

\(C=2.\frac{2}{3}.\left(3.\frac{2}{3}+1\right)=2.\frac{2}{3}.3=4\)

Vậy C = 4 với \(x=\frac{2}{3}\)

(4x² - 4x + 1) - (x + 1)² 

 =(2x - 1)² - (x + 1)² 

 = (2x - 1 + x + 1)(2x - 1 - x - 1) 

= 3x(x - 2) 

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

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

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

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

\(VT\ge0\Rightarrow VP\ge0\Rightarrow x^3-1\ge0\Leftrightarrow x\ge1\)

Với \(x\ge1\)thì \(\left|x+1\right|=x+1,\left|x^2+x-2\right|=x^2+x-2\)

Phương trình ban đầu tương đương với:

\(x+1+x^2+x-2=x^3-1\)

\(\Leftrightarrow x^3-x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=2\)(vì \(x\ge1\))