Đáp án:
a, x.(2x+1) ²- x ². (x+2)+x ³-x+3
= x.(2x+1)(2x+1)-1x²(x+2)+x³+3
= 𝑥(2x(2𝑥+1)+1(2x+1))−1𝑥²(𝑥+2)+𝑥³−𝑥+3
=x(4x²+2x+1(2𝑥+1))−1𝑥²(𝑥+2)+𝑥³−𝑥+3
=𝑥(4𝑥²+2𝑥+2𝑥+1)−1𝑥²(𝑥+2)+𝑥³−𝑥+3
=𝑥(4𝑥²+4x+1)−1𝑥²(𝑥+2)+𝑥³−𝑥+3
=4x³+4x²+x−1𝑥²(𝑥+2)+𝑥³−𝑥+3
=4𝑥3+4𝑥2+𝑥−(x³+2x²)+𝑥3−𝑥+3
=4𝑥3+4𝑥2+𝑥−𝑥3−2𝑥2+𝑥3−𝑥+3
=4𝑥³+2x²+𝑥−𝑥3+𝑥3−𝑥+3
=5x³+2𝑥²+𝑥−𝑥³−𝑥+3
=5𝑥³+2𝑥²-x³+3
=4x³+2x²+3
b,(x-1).(x+1).(x-4)-x ³-4
=x(x-4)(x+1)-1(x-4)(x+1)-1x³-4
=(x+1).x²-4x(x+1)−1(𝑥−4)(𝑥+1)−1𝑥³−4
=x³+1x²−4𝑥(𝑥+1)−1(𝑥−4)(𝑥+1)−1𝑥³−4
=𝑥³+𝑥²−4𝑥(𝑥+1)−1(𝑥−4)(𝑥+1)−1𝑥³−4
=x³+x²-4x²-4x−1(𝑥−4)(𝑥+1)−1𝑥³−4
=𝑥³−3𝑥²−4𝑥−1(𝑥−4)(𝑥+1)−1𝑥³−4
=𝑥³−3𝑥²−4𝑥−(𝑥(𝑥+1)−4(𝑥+1))−1𝑥³−4
=𝑥³−3𝑥²−4𝑥−(𝑥²+𝑥−4(𝑥+1))−1𝑥³−4
=𝑥³−3𝑥²−4𝑥−(𝑥²+𝑥−4𝑥−4)−1𝑥³−4
=𝑥³−3𝑥²−4𝑥−(𝑥²−3𝑥−4)−1𝑥³−4
=𝑥³−3𝑥²−4𝑥−𝑥²+3𝑥+4−1𝑥³−4
=𝑥³−3𝑥²−𝑥−𝑥²+4−1𝑥³−4
=𝑥³−3𝑥²−𝑥²−𝑥+4−1𝑥³−4
=𝑥³−3𝑥²−𝑥²−𝑥−𝑥³
=-4x²-x
c,(x+y+z+t).(x+y-z-t)-
=x²+xy-xz-xt+xy+y²-yz-yt+xz+yz-z²-zt+xt+yt-zt-t²-x ²-y ²+z ²+t ²
=(x²+y²-z²-t²)-(x ²+y ²-z ²-t ²)+2xy-2zt
=2xy-2zt
