W tym zadaniu skorzystamy ze wzoru na:
- kwadrat sumy:
(a+b)2=a2+2ab+b2
- kwadrat różnicy:
(a−b)2=a2−2ab+b2
- różnicę kwadratów:
a2−b2=(a−b)(a+b)
a) (1+2)2+(1−2)2=(1+22+2)+(1−22+2)=
=(3+22)+(3−22)=3+22+3−22=6
b) (3−1)2−(2−3)2=(3−23+1)−(4−43+3)=
=(4−23)−(7−43)=4−23−7+43=23−3
c) (23−23)2−(23+23)2=
=[(23−23)−(23+23)]⋅[(23−23)+(23+23)]=
=[23−23−23−23]⋅[23−23+23+23]=
=−26⋅43=−3⋅43=−123
d) (31+32)2−(31−32)2=
=91+2⋅31⋅32+(32)2−(91−2⋅31⋅32+(32)2)=
=91+22+9⋅2−91+22−9⋅2=42
e) (4−5)(4+5)−(5−2)(2+5)=
=42−52−(5−2)(5+2)=
=16−5−(52−22)=11−(5−4)=11−1=10
f) (6−5)(6+5)+(6−5)2=
=(62−52)+(6−26⋅5+5)=
=(6−5)+(11−230)=1+11−230=12−230
g) (25−10)2−(25+1)(1−25)=
=(4⋅5−45⋅10+10)−(1+25)(1−25)=
=20−450+10−(12−(25)2)=30−450−(1−4⋅5)=
=30−4⋅25⋅2−1+20=30−4⋅5⋅2+19=49−202
h) (6−23)2−(52−1)(1+52)=
=62−2⋅6⋅23+(23)2−((52)2−12)=
=6−418+4⋅3−(25⋅2−1)=6−4⋅32+12−(50−1)=
=6−122+12−49=−31−122
