\(A=\frac{2x^2+4x}{x^3-4x}+\frac{x^2-4}{x^2+2x}+\frac{2}{2-x}\left(x\ne0;x\ne\pm2\right)\)

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

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

\(A=\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}+\frac{x^3-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}-\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{2x^2+4x+x^3-2x^2-4x+8-2x^2-4x}{x\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x\left(x+2\right)+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x+8}{x\left(x-2\right)}\)

Vậy \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)

b) \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)

Ta có: x=4 (tmđk) thay vào A ta có:

\(A=\frac{-2\cdot4+8}{4\left(4-2\right)}=\frac{-8+8}{4\cdot2}=\frac{0}{8}=0\)

Vậy A=0 với x=4

2018^2 - 2017^2 = 435

2018^2 = 4072324

2017^2 = 4068289

Vậy 2018^2 - 2017^2 = 4072324 - 4068289 = 4035

\(\text{x+3=-2x-4+5x}\)

\(x+2x-5x=-4-3\)

\(-2x=-7\)

\(x=-7:\left(-2\right)\)

\(x=\frac{7}{2}\)

học tốt

\(x^2\left(x-1\right)+16\left(1-x\right)\)

\(=x^2\left(x-1\right)-16\left(x-1\right)\)

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

\(=\left(x-1\right)\left(x-4\right)\left(x-4\right)\)

=x²(x-1)-16(x-1)

=(x-1) (x²-16)

=(x-1)(x-4)(x+4)