Đáp án:
Giải thích các bước giải:
Bài 3
3) Ta có
x(2x2−3)−x2(5x+1)+x2=2x3−3x−5x3−x2+x2x(2x2−3)−x2(5x+1)+x2=2x3−3x−5x3−x2+x2
=−3x3−3x=−3x(x2+1)=−3x3−3x=−3x(x2+1)
4) Ta có
3x(x−2)−5x(x−1)−8(x2−3)=3x2−6x−5x2+5x−8x2+243x(x−2)−5x(x−1)−8(x2−3)=3x2−6x−5x2+5x−8x2+24
=−10x2−x+24=−10x2−x+24
Bài 4
a) Ta có
1012=(100+1)2=1002+2.100+1=10000+200+1=102011012=(100+1)2=1002+2.100+1=10000+200+1=10201
b) Ta có
97.103=(100−3)(100+3)=1002−32=10000−9=999197.103=(100−3)(100+3)=1002−32=10000−9=9991
c) Ta có
772+232+77.46=772+232+77.23.2772+232+77.46=772+232+77.23.2
=(77+23)2=(77+23)2
=1002=10000=1002=10000
d) Ta có
1052−52=(105−5)(105+5)=100.110=11000
