BT 3: tHỰC HIỆN PHÉP TÍNH
a)(3x+7)(2x+3)−(3x−5)(2x+11)a)(3x+7)(2x+3)-(3x-5)(2x+11)
=6x2+9x+14x+21−(6x2+33x−10x−55)=6x2+9x+14x+21-(6x2+33x-10x-55)
=6x2+9x+14x+21−6x2−33x+10x+55=6x2+9x+14x+21-6x2-33x+10x+55
=(6x2−6x2)+(9x+14x−33x+10x)+(21+55)=(6x2-6x2)+(9x+14x-33x+10x)+(21+55)
= 76= 76
b)(x−1)2+(x+1)2−(x+3)(x−3)b)(x-1)2+(x+1)2-(x+3)(x-3)
=x2−2x+1+x2+2x+1−x2+9=x2-2x+1+x2+2x+1-x2+9
=(x2+x2−x2)−(2x−2x)+(1+1+9)=(x2+x2-x2)-(2x-2x)+(1+1+9)
=x2+11=x2+11
c)(x+2)2+x(x−x2)−(x+1)(7x+5)c)(x+2)2+x(x-x2)-(x+1)(7x+5)
=x2+4x+4+x2−x3−7x2−5x+7x−5=x2+4x+4+x2-x3-7x2-5x+7x-5
=−x3+(x2+x2−7x2)+(4x−5x−7x)+(4−5)=-x3+(x2+x2-7x2)+(4x-5x-7x)+(4-5)
=−x3−5x2−8x−1=-x3-5x2-8x-1
d)(3x+2)(9x2−6x+4)−9x(3x2+1)+9xd)(3x+2)(9x2-6x+4)-9x(3x2+1)+9x
=27x3−18x2+12x+18x2−12x+8−27x3−9x+9x=27x3-18x2+12x+18x2-12x+8-27x3-9x+9x
=(27x3−27x3)−(18x2−18x2)+(12x−12x−9x+9x)+8=(27x3-27x3)-(18x2-18x2)+(12x-12x-9x+9x)+8
=8=8
e)(2x+3)2+(2x+5)2−2(2x+3)(2x+5)e)(2x+3)2+(2x+5)2-2(2x+3)(2x+5)
=[(2x+3)−(2x+5)]2=[(2x+3)-(2x+5)]2
=(2x+3−2x−5)2=(2x+3-2x-5)2
=(−2)2=(-2)2
=4=4
f)(3x−1)2+2(9x2−1)+(3x+1)2f)(3x-1)2+2(9x2-1)+(3x+1)2
=(3x−1)2+2(3x−1)(3x+1)+(3x+1)2=(3x-1)2+2(3x-1)(3x+1)+(3x+1)2
=[(3x−1)+(3x+1)]2=[(3x-1)+(3x+1)]2
=(3x−1+3x+1)2=(3x-1+3x+1)2
=(6x)2=(6x)2
=36x2=36x2
