Đáp án:
b1
a,
b,
=𝑥^2+𝑥𝑦+𝑦(𝑥+𝑦)−1(𝑥−𝑦)^2−4(𝑥−1)⋅𝑦
=𝑥^2+2𝑥𝑦+𝑦^2−1(𝑥−𝑦)^2−4(𝑥−1)⋅𝑦
=𝑥^2+2𝑥𝑦+𝑦^2−1(𝑥−𝑦)(𝑥−𝑦)−4(𝑥−1)⋅𝑦
=𝑥^2+2𝑥𝑦+𝑦^2−1(𝑥^2−𝑥𝑦−𝑦(𝑥−𝑦))−4(𝑥−1)⋅𝑦
=𝑥^2+2𝑥𝑦+𝑦^2−1(𝑥^2−𝑥𝑦−𝑥𝑦+𝑦^2)−4(𝑥−1)⋅𝑦
=𝑥^2+2𝑥𝑦+𝑦^2−1(𝑥^2−2𝑥𝑦+𝑦^2)−4(𝑥−1)⋅𝑦
=𝑥^2+2𝑥𝑦+𝑦^2−1𝑥^2+2𝑥𝑦−1𝑦^2−4𝑥𝑦+4𝑦
=2𝑥𝑦+𝑦^2+2𝑥𝑦−1𝑦^2−4𝑥𝑦+4𝑦
=𝑦^2−1𝑦^2+4𝑦
=4y
b2
a,=x^2+xy+3(x+y)
=x(x+y)+3(x+y)
=(x+3)(x+y)
b,x^3+5x^2+6x
=x(x^2+3x+2x+6)
=x(x+2)(x+3)
b3
a, (x+3)/(2x+1)-(x-7)(2x+1)
=(x+3-x+7)/(2x+1)
=10/(2x+1)
b,