第一章 整式的乘除 训练2023-2024学年北师大版七年级数学下册

整式的乘除运算 1、 整式的乘法运算 (1)3a(2a²-9a+3)-4a(2a-1) (2)x²(3-x)+x(x²-2x)+1 (3) (-3x²y)(-2xy+3yz+1) (4)(-4x)(2x²+3x-1) (5)(x+y)(x²-xy+y²) (6)(x-2y)( x+4y)-(2x-y)(x+y) (7)(a+b)(a-b)+2a² (8)(a+2b)(a-2b)+2(a-b)² (9) (a+2)²-(1-a)(-a-1) (10)(2x-y)(3x-2y)+(x-3y)(x+3y)-(2x-3y)² (11)(x-3y)(x+6y)-(3x+y)( x-y) (12)(a-b-c)·(a+b-c) (13)(x+y-1)(x+y+1)-(x-2y)(x+2y) (14)(a-b)(a+b)(a²+b²)(a⁴+b⁴)·……·(+) (15) (x-)²(x²+)²(x+)² (16)(2a+b)(2a-b)-(2b-3a)(3a+2b) (17) (2x+3y)²(2x-3y)² (18)(3x-1)²+(1+3x)(1-3x) (19)(a-2b+c)(a+2b-c) (20)(m+n)(x-y)-(m+n)(x+y) 2、整式的除法 (1)-30a²b²C÷6a²b² (2)100c²÷(-5a²b)² (3)(4x³y)³÷y² (4)(14x³+4x²-2x)÷2x (5)()÷(ab²)² (6)[a²(a+b)-3a²b]÷2a (7) xy²(3x+4y)÷2xy (8)[(x+y)(x-2y)-x²]÷(-2y) (9)[ 6y²-(-3x+2y)(2x+3y)]÷3x (10)(y)³÷(-x²y)²+(-x²y)³·x²÷y² 3、易错题型 ( 1) 计算：(x+y)⁵÷(-x-y)²÷(x+y) (2) 若,求2m-6n的值. (3) 计算 ……. (4) 试说明 (n+7)²-(n-5)² (n为正整数)能被24整除. (5)已知 5x²-x-1=0, 求代数式(3x+2)(3x-2)+x(x-2)的值. (6)若a+b=-3,ab=1,则 (a+1)(b+1)(a-1)(b-1)=____________. (7) 已知多项式 x²+ax+1 与x-b的乘积中含x²的项的系数为3,含x项的系数为-9, 求 a²+b²的值. 答案： 1. (1)6a³-35a²+13a (2) x²+1 (3)6x³y²-9x²y²z-3x²y (4)-8x³-12x²+4x (5)x³+y³ (6)-x²+xy-7y² (7) 3a²-b² (8)3a²-4ab-2b² (9) 4a+5 (10)3x²+5xy-16y² (11)-2x²+5xy-17y² (12) a²-2ac+c²-b² (3) 5y²+2xy-1 (14) (15)x⁸－x⁴－ (16)13a²-5b² (17)16x⁴-72x²y²+81y⁴ (18)-6x+2 (19) a²-4b²+4bc-c² (20)-2my-2ny 2.(1)-5c (2) 4ab³c² (3)64y (4) 7x²+2x-1 (5)a³b⁴-8b² (6)a²-ab (7)xy+y² (8)x+y (9)2x+y (10) 0 3.易错题型 (1)原式 =÷(x+y)²÷(x+y) = =(x+y)² =x²+2xy+y² (2) 变形 =2² ∴m-3n=2 2m-6n=2(m-3n)=2×2=4 (3)原式: …… =…… = = (4) (n+7)²-(n-5)² =n²+14n+49-(n²-10n+25) =24n+24 =24(n+1) ∵n为正整数,所以24(n+1)能被24整除 即 (n+7)²-(n-5)² 能被24整除 (5)代数式 =9x²-4+x²-2x =10x²-2x-4 =2( 5x²-x)-4 因为 5x²-x-1=0 所以5x²-x=1 ∴ 2(5n²-x)-4=2×1-4=-2 (6)答案：-5 (a+1)(b+1)(a-1)(b-1)=(a²-1)(b²-1)=a²b²-a²-b²+1 =(ab)²-(a²+b²)+1 :a+b=-3,ab=1 ∴a²+b²=(-3)²-2×1=7 ∴(ab)²-(a²+b²)+1=1-7+1=-5 (7)(x²+ax+1)(x-b) =x³-bx²+ax²-abx+x-b =x³+(a-b)x