(1)(2ab+3b2﹣5)﹣(3ab+3b2﹣8)(2)x2×x3+x7×x2(3)107÷(103÷102)(4)(x﹣y)3×(x﹣y)2×(y﹣x)
题型:解答题难度:一般来源:云南省月考题
(1)(2ab+3b2﹣5)﹣(3ab+3b2﹣8) (2)x2×x3+x7×x2 (3)107÷(103÷102) (4)(x﹣y)3×(x﹣y)2×(y﹣x) |
答案
解:(1)(2ab+3b2﹣5)﹣(3ab+3b2﹣8) =2ab+3b2﹣5﹣3ab﹣3b2+8 =(2ab﹣3ab)+(3b2﹣3b2)+8 =﹣ab+8; (2)x2×x3+x7÷x2 =x2+3+x7﹣2 =x5+x5 =2x5; (3)107÷(103÷102) =107÷103﹣2 =107÷10 =107﹣1 =106; (4)(x﹣y)3×(x﹣y)2×(y﹣x) =﹣(x﹣y)3×(x﹣y)2×(x﹣y) =﹣(x﹣y)3+2+1 =﹣(x﹣y)6. |
举一反三
(1)(2x2)3﹣6x3(x3+2x2+x) (2)x(x﹣y)+(2x+y)(x﹣y) (3)(3mn+1)(3mn﹣1)﹣8m2n2 (4)[(x+y)2﹣(x﹣y)2]÷(2xy) |
求值:x(x+2y)﹣(x+1)2+2x,其中. |
若a的值使得x2+4x+a=(x+2)2﹣1成立,则a的值为 |
[ ] |
A.5 B.4 C.3 D.2 |
已知2x﹣y=10,求代数式[(x2+y2)﹣(x﹣y)2+2y(x﹣y)]÷4y的值. |
化简求值:[(x+2y)2﹣(x+y)(3x﹣y)﹣5y2]÷2x,其中x=﹣2,y=. |
最新试题
热门考点