1、3计算(c2)n(cn+1)2等于(Ac4n+2BcCcDc3n+44与( 2a2)35的值相等的是(A 25a30 B 215a 30C( 2a2)15D( 2a)305下列计算正确的是(A(xy)3 = xy3 B(2xy)3 = 6x3y3C(3x2)3 = 27x5D(a2b)n = a2nbn6下列各式错误的是()A(23)4 = 212 B( 2a)3 = 8a3C(2mn2)4 = 16m4n8 D(3ab)2 = 6a2b27下列各式计算中,错误的是(A(m6)6 = m36 B(a4)m = (a 2m)2Cx2n = (xn)2 Dx2n = (x2)n二、解答题:1已知3
2、2n+1+32n = 324,试求n的值2已知 2m = 3,4n = 2,8k = 5,求 8m+2n+k的值3计算:x2(x3)24 4如果am = 5,an = 7,求a 2m+n的值幂的运算测试题答案:1、D说明:mnm2n+1 = mn+2n+1 = m3n+1,A中计算正确;(am1)2 = a2(m1) = a 2m2,B中计算正确; (a2b)n = (a2)nbn = a2nbn,C中计算正确;(3x2)3 = (3)3(x2)3 = 27x6,D中计算错误;所以答案为D2、B因为xa = 3,xb = 5,所以xa+b = xaxb = 35 = 15,答案为B3、A(c2
3、)n(cn+1)2 = c2nc2(n+1) = c2nc2n+2 = c2n+2n+2 = c4n+2,所以答案为A4、C( 2a2)35 = ( 2a2)35 = ( 2a2)15,所以答案为C5、D(xy)3 = x3y3,A错;(2xy)3 = 23x3y3 = 8x3y3,B错;(3x2)3 = (3)3(x2)3 = 27x6,C错;(a2b)n = (a2)nbn = a2nbn,D正确,答案为D6、C(23)4 = 234 = 212,A中式子正确;( 2a)3 = (2) 3a3 = 8a3,B中式子正确;(3ab)2 = 32a2b2 = 9a2b2,C中式子错误;(2mn
4、2)4 = 24m4(n2)4 = 16m4n8,D中式子正确,所以答案为C7、D(m6)6 = m66 = m36,A计算正确;(a4)m = a 4m,(a 2m)2 = a 4m,B计算正确;(xn)2 = x2n,C计算正确;当n为偶数时,(x2)n = (x2)n = x2n;当n为奇数时,(x2)n = x2n,所以D不正确,答案为D1解:由32n+1+32n = 324得332n+32n = 324,即432n = 324,32n = 81 = 34,2n = 4,n = 22解析:因为 2m = 3,4n = 2,8k = 5所以 8m+2n+k = 8m82n8k = (23
5、)m(82)n8k = 23m(43)n8k = ( 2m)3(4n)38k = 33235 = 2785 = 10803答案:x32解:x2(x3)24 = (x2x32)4 = (x2x6)4 = (x2+6)4= (x8)4 = x84 = x324答案:a 2m+n = 175因为am = 5,an = 7,所以a 2m+n = a 2man = (am)2an = (5)27 = 257 = 175第二单元 整式的乘法测试题1对于式子(x2)n xn+3(x0),以下判断正确的是(Ax0时其值为正 Bxx35的解集为x9(x2)(x+3)的正整数解6计算:3y(y4)(2y+1)(2
6、y3)(4y2+6y9)整式的乘法测试题答案:1 C(x2)n的符号由n的奇偶性决定当n为奇数时,n+1为偶数,则只要x0,xn+1即为正,所以(x2) n xn+3 = (xn+1)3,为正;n为偶数时,n+1为奇数,则xn+1的正负性要由x的正负性决定,因此(x2) n xn+3 = (xn+1)3,其正负性由x的正负性决定;所以正确答案为C2 D(xyz)2(yx+z)(zx+y) = (xyz)4,因此,代数式(xyz)2(yx+z)(zx+y)的值一定是非负数,即正确答案为D3 B原方程变形为:x23x23x = 5x2x2+8,8x = 8,x = 1,答案为B4 C利用长方体的体
7、积公式可知该长方体的体积应该是长宽高,即( 3a4) 2aa = 6a3 8a2,答案为C5 D(a4+ 4a2+16) a24(a4+ 4a2+16) = a6+ 4a4+ 16a2 4a4 16a264 = (2)664 = 0,答案为D6 A(x3y+4z)(6x) = 6x2+18xy24xz,A错,经计算B、C、D都是正确的,答案为A7 A(3x4y)(5x+6y) = 15x2+18xy20xy24y2 = 15x22xy24y2,A错;经计算B、C、D都正确,答案为A8 D(6ab2 4a2b)3ab = 6ab23ab 4a2b3ab = 18a2b3 12a3b,A计算错误;
8、(x)(2x+x21) = x2x+(x)x2(x) = 2x2x3+x = x32x2+x,B计算错误;(3x2y)(2xy+3yz1) = (3x2y) (2xy)+(3x2y) 3yz(3x2y) = 6x3y29x2y2z+3x2y,C计算错误;(a3b)2ab = (a3) 2ab(b)2ab =a4bab2,D计算正确,所以答案为D9 B因为(x2)(x+3) = xx2x+3x6 = x2+x6,所以a = 1,b = 6,答案为B10 D( 2a1)( 5a+2) = 2a 5a1 5a+ 2a212 = 10a2 5a+ 4a2 = 10a2a2,所以答案为D1 2003(3x2)(x22x3)+3x(x32x23x)+2003 = 3x4+6x3+9x2+3x46x39x2+2003 = 20032 x =将原方程化简,6x213x+6 = 6x2x5,12x = 11,x =3原式= 6y2+18y+18 = 25
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