= \(\dfrac{2+x}{2-x}\)- \(\dfrac{4x^2}{^{^{ }}4-x^2}\)- \(\dfrac{2-x}{2+x}\): \(\dfrac{x^2-3x}{2x^2-x^3}\)
= \(\dfrac{\left(2+x\right)\left(2+x\right)-4x^{2^{ }}+\left(2-x\right)\left(2-x\right)}{4-x^2}\) . \(\dfrac{2x^{2^{ }}-x^3}{x^{2^{ }}-3x}\)
= \(\dfrac{4+4x+x^{2^{ }}-4x^{2^{ }}-4+4x-x^2}{^{ }4-x^2}\). \(\dfrac{2x^{2^{ }}-x^3}{x^2-3x}\)
= \(\dfrac{8x-4x^2}{4-x^2}\). \(\dfrac{2x^2-x^3}{x^2-3x}\)
= \(\dfrac{8-4x^2}{2+x}\). \(\dfrac{x^2}{x-3}\)
mình viết nhầm hàng thứ 2 phải là -(2-x)(2-x)
= 2+x2−x2+x2−x- 4x24−x24x24−x2- 2−x2+x2−x2+x: x2−3x2x2−x3x2−3x2x2−x3
= (2+x)(2+x)−4x2+(2−x)(2−x)4−x2(2+x)(2+x)−4x2+(2−x)(2−x)4−x2 . 2x2−x3x2−3x2x2−x3x2−3x
= 4+4x+x2−4x2−4+4x−x24−x24+4x+x2−4x2−4+4x−x24−x2. 2x2−x3x2−3x2x2−x3x2−3x
= 8x−4x24−x28x−4x24−x2. 2x2−x3x2−3x2x2−x3x2−3x
= 8−4x22+x8−4x22+x. x2x−3
