∴(x﹣)2=4, ∴x2+∴x2+
﹣2=4, =6;
(2)∵x+y=3,x2+y2=5, ∴(x+y)2=9, ∴x2+2xy+y2=9, ∴2xy=9﹣(x2+y2), ∴2xy=4, ∴2(x﹣y)2 =2(x2+y2﹣2xy) =2×(5﹣4) =2×1 =2.
25.【解答】解:(1)等式右边=(a2﹣2ab+b2+b2﹣2bc+c2+a2﹣2ac+c2) =(2a2+2b2+2c2﹣2ab﹣2bc﹣2ac) =a2+b2+c2﹣ab﹣bc﹣ac=等式左边,
则a2+b2+c2﹣ab﹣bc﹣ac=[(a﹣b)2+(b﹣c)2+(a﹣c)2];
(2)由a=2019,b=2020,c=2021,得到a﹣b=﹣1,a﹣b=﹣2,b﹣c=﹣1, 则a2+b2+c2﹣ab﹣bc﹣ac
=[(a﹣b)2+(b﹣c)2+(a﹣c)2] =×(1+4+1) =3;
(3)∵a﹣b=,b﹣c=, ∴a﹣c=,
∵a2+b2+c2=1,a2+b2+c2﹣ab﹣bc﹣ac=[(a﹣b)2+(b﹣c)2+(a﹣c)2], ∴1﹣(ab+bc+ac)=×(
+
+
)
第9页(共10页)
则ab+bc+ac=1﹣=﹣.
26.【解答】解:(1)根据题意得:S1=a(a+4)=a2+4a,S2=(a+2)2=a2+4a+4, ∵S1﹣S2=(a2+4a)﹣(a2+4a+4)=a2+4a﹣a2﹣4a﹣4=﹣4<0, ∴S1<S2;
故答案为:a2+4a,a2+4a+4;
(2)∵A=2a2﹣6a+1,B=a2﹣4a﹣1,
∴A﹣B=2a2﹣6a+1﹣a2+4a+1=a2﹣2a+2=a2﹣2a+1+1=(a﹣1)2+1≥1>0, 则A>B;
(3)由M=N,得到M﹣N=0, ∴(a﹣4)2﹣16+(a﹣6)2=0,
整理得:a2﹣10a+18=0,即a2﹣10a=﹣18, 则(a﹣4)(a﹣6)=a2﹣10a+24=﹣18+24=6.
第10页(共10页)
相关推荐: