Dạng 2. Chứng minh đẳng thức có đáp án

$\frac{x^{2}y+2x{y}^2+y^3}{2x^2+xy-y^2}=\frac{y\left(x^2+2xy+y^2\right)}{2x^2+2xy-xy-y^2}=\frac{y{\left(x+y\right)}^2}{2x\left(x+y\right)-y\left(x+y\right)}$

$=\frac{y{\left(x+y\right)}^2}{\left(x+y\right)\left(2x-y\right)}=\frac{y\left(x+y\right)}{2x-y}=\frac{xy+y^2}{2x-y}$

$\frac{x^2+3xy+2y^2}{x^3+2x^{2}y-x{y}^2-2y^3}=\frac{x^2+xy+2xy+2y^2}{x^3-x{y}^2+2x^{2}y-2y^3}=\frac{x\left(x+y\right)+2y\left(x+y\right)}{x\left(x^2-y^2\right)+2y\left(x^2-y^2\right)}$

$=\frac{\left(x+y\right)\left(x+2y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}=\frac{\left(x+y\right)\left(x+2y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}$

$\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{a^3-a-4a^2+4}{a^3-8-7a^2+14a}=\frac{a\left(a^2-1\right)-4\left(a^2-1\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}$

$=\frac{\left(a^2-1\right)\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4-7a\right)}=\frac{\left(a^2-1\right)\left(a-4\right)}{\left(a-2\right)\left(a^2-5a+4\right)}=\frac{\left(a-1\right)\left(a+1\right)\left(a-4\right)}{\left(a-2\right)\left(a-4\right)\left(a-1\right)}=\frac{a+1}{a-2}$

$\frac{x^2{y}^2+1+\left(x^2-y\right)\left(1-y\right)}{x^2{y}^2+1+\left(x^2+y\right)\left(1+y\right)}=\frac{x^2{y}^2+1+x^2-x^{2}y-y+y^2}{x^2{y}^2+1+x^2+x^{2}y+y+y^2}=\frac{x^2{y}^2-x^{2}y+x^2+y^2-y+1}{x^2{y}^2+x^{2}y+x^2+y^2+y+1}$

$=\frac{\left(y^2-y+1\right)\left(x^2+1\right)}{\left(y^2+y+1\right)\left(x^2+1\right)}=\frac{y^2-y+1}{y^2+y+1}$