24.已知a2+b2+c2=ab+bc+ac,求证a=b=c.
[答案]用配方法,a2+b2+c2-ab-bc-ac=0,∴ 2(a2+b2+c2-ab-ac-bc)=0,
即(a-b)2+(b-c)2+(c-a)2=0.∴ a=b=c.
23.已知(a+b)2=10,(a-b)2=2,求a2+b2,ab的值.
[答案]a2+b2=[(a+b)2+(a-b)2]=6,
ab=[(a+b)2+(a-b)2]=2.
22.已知a+b=5,ab=7,求,a2-ab+b2的值.
[答案]=[(a+b)2-2ab]=(a+b)2-ab=.
a2-ab+b2=(a+b)2-3ab=4.
21.已知a2+6a+b2-10b+34=0,求代数式(2a+b)(3a-2b)+4ab的值.
[提示]配方:(a+3)2+(b-5)2=0,a=-3,b=5,
[答案]-41.
20.用简便方法计算:(每小题3分,共9分)
(1)982;
[答案](100-2)2=9604.
(2)899×901+1;
[答案](900-1)(900+1)+1=9002=810000.
(3)()2002·(0.49)1000.
[答案]()2·()2000·(0.7)2000=.
19.(1)(-3xy2)3·(x3y)2; [答案]-x9y8.
(2)4a2x2·(-a4x3y3)÷(-a5xy2);[答案]ax4y.
(3)(2a-3b)2(2a+3b)2;[答案]16a4-72a2b2+81b4.
(4)(2x+5y)(2x-5y)(-4x2-25y2); [答案]625y4-16x4.
(5)(20an-2bn-14an-1bn+1+8a2nb)÷(-2an-3b);[答案]-10abn-1+7a2bn-4an+3.
(6)(x-3)(2x+1)-3(2x-1)2.
[答案]-10x2+7x-6.
18.如果x2-kx-ab=(x-a)(x+b),则k应为…………………………………( )
(A)a+b (B)a-b (C)b-a (D)-a-b
[答案]B.
17.下列各式中正确的是………………………………………………………………( )
(A)(a+4)(a-4)=a2-4 (B)(5x-1)(1-5x)=25x2-1
(C)(-3x+2)2=4-12x+9x2 (D)(x-3)(x-9)=x2-27
[答案]C.
16.下列各组数中,互为相反数的是……………………………………………………( )
(A)(-2)-3与23 (B)(-2)-2与2-2
(C)-33与(-)3 (D)(-3)-3与()3
[答案]D.
15.若a≠b,下列各式中不能成立的是………………………………………………( )
(A)(a+b)2=(-a-b)2 (B)(a+b)(a-b)=(b+a)(b-a)
(C)(a-b)2n=(b-a)2n (D)(a-b)3=(b-a)3