giải các phương trình sau :
a. (x-3)(x-4)-2.(3x-2)=\(\left(4-x\right)^2\)
b. \(\left(x+2\right)\left(x-2\right)+5x^2=\left(3x+1\right)-3x^2\)
c. \(\left(x+2\right)^3-\left(x-1\right)^3=\left(3x+1\right).\left(3x-1\right)\)
d.\(\frac{3-x}{2018}+\frac{x-1}{2020}=\frac{-x}{2021}+1\)
a) Ta có: \(\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)=\left(4-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)-2\left(3x-2\right)-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)-\left(x-4\right)\right]-2\left(3x-2\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3-x+4\right)-6x+4=0\)
\(\Leftrightarrow x-4-6x+4=0\)
\(\Leftrightarrow-5x=0\)
mà -5<0
nên x=0
Vậy: x=0