Đề:............

<=> - (1 - 2018x) + 2019x.(1 - 2018x) = 0

<=> (1 - 2018x).[(-1) + 2019x] = 0

Xét 2 trường hợp, ta có:

TH₁: 1 - 2018x = 0          TH₂: -1 + 2019x = 0

<=> 2018x = 1                 <=> 2019x = 1

<=> x = 1/2018                <=> x = 1/2019

Vậy x = 1/2018; 1/2019

Bài 1 : 

\(49\left(x-2\right)^2-25\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[7\left(x-2\right)-5\left(2x+1\right)\right]\left[7\left(x-2\right)+5\left(2x+1\right)\right]=0\)

\(\Leftrightarrow\left(7x-14-10x-5\right)\left(7x-14+10x+5\right)=0\)

\(\Leftrightarrow\left(-3x-19\right)\left(17x-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-3x=19\\17x=9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-19}{3}\\x=\frac{9}{17}\end{cases}}}\)

Bài 2 : 

+) \(9x^2-6xy+y^2-21x+7y\)

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

\(=\left(3x-y\right)\left(3x-y-7\right)\)

+) \(x^2+2x-35\)

\(=x^2+2x+1-36\)

\(=\left(x+1-6\right)\left(x+1+6\right)\)

\(=\left(x-5\right)\left(x+7\right)\)

+) \(2x^2+9x-5\)

\(=2x^2-x+10x-5\)

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

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

+) \(6x^2+23x+15\)

\(=6x^2+18x+5x+15\)

\(=6x\left(x+3\right)+5\left(x+3\right)\)

\(=\left(x+3\right)\left(6x+5\right)\)

\(a=x^3-6x^2+12x-8-x^3+4x^2+8\)

\(a=-2x^2+12x\)

\(a=\left(x-2\right)^3-x^2\left(x-4\right)+8\)

\(a=x^3-6x^2+12x-8-x^3+4x^2+8\)

\(a=-2x^2+12x\)

\(16\left(2x+3\right)^2-9\left(5x-2\right)^2\)

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

\(=\left(8x+12\right)^2-\left(15x-6\right)^2\)

\(=\left[\left(8x+12\right)-\left(15x-6\right)\right].\left[\left(8x+12\right)+\left(15x-6\right)\right]\)

\(=\left(8x+12-15x+6\right)\left(8x+12+15x-6\right)\)

\(=\left(-7x+18\right)\left(23x+6\right)\)