内容正文:
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湖北世纪华章文化传播有限公司
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第一章 整式的乘除
小专题2 整式的乘除运算
数 学
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类型1 整式的运算
1.计算:
(1)(-x)5÷(-x)-2÷(-x)3;
解:原式=(-x)5-(-2)-3
=x4.
(2)(-5a2b4c2)2÷(-ab2c)3;
解:原式=25a4b8c4÷(-a3b6c3)
=-25ab2c.
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(3)(-3x2y)2·(-eq \f(2,3)xyz)·eq \f(3,4)xz2;
解:原式=9x4y2·(-eq \f(2,3)xyz)·eq \f(3,4)xz2
=-eq \f(9,2)x6y3z3.
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(4)(2020·武汉)[a3·a5+(3a4)2]÷a2.
解:原式=(a8+9a8)÷a2
=10a8÷a2
=10a6.
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2.计算:
(1)(3x-1)(2x+1);
解:原式=6x2+3x-2x-1
=6x2+x-1.
(2)(x-1)(x2+x+1);
解:原式=x3+x2+x-x2-x-1
=x3-1.
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(3)(-2a2)·(3ab2-5ab3)+8a3b2;
解:原式=-6a3b2+10a3b3+8a3b2
=2a3b2+10a3b3.
(4)(a+1)2+2(1-a);
解:原式=a2+2a+1+2-2a
=a2+3.
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(5)(y+2)(y-2)-(y-1)(y+3).
解:原式=y2-4-(y2+3y-y-3)
=-2y-1.
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3.计算:
(1)(3x2y-xy2+eq \f(1,2)xy)÷(-eq \f(1,2)xy);
解:原式=-6x+2y-1.
(2)(3x2)2·(-4y3)÷(6xy)2;
解:原式=9x4·(-4y3)÷36x2y2
=-36x4y3÷36x2y2
=-x2y.
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(3)[(2x-y)(2x+y)+y(y-6x)]÷2x;
解:原式=(4x2-y2+y2-6xy)÷2x
=(4x2-6xy)÷2x
=2x-3y.
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(4)(a+b+c)(a-b+c).
解:原式=[(a+c)+b][(a+c)-b]
=(a+c)2-b2
=a2+2ac+c2-b2.
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类型2 利用直接代入进行化简求值
4.先化简,再求值:
(1)(2020·常州)(x+1)2-x(x+1),其中x=2;
解:原式=x2+2x+1-x2-x
=x+1.
当x=2时,原式=2+1=3.
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(2)(a+b)(a-2b)-(a+2b)(a-b),其中a=-2,b=eq \f(2,3);
解:原式=a2-ab-2b2-(a2+ab-2b2)
=a2-ab-2b2-a2-ab+2b2
=-2ab.
当a=-2,b=eq \f(2,3)时,原式=-2×(-2)×eq \f(2,3)=eq \f(8,3).
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(3)(x+y)(x-y)-(4x3y-8xy3)÷2xy,其中x=-1,y=eq \f(1,2);
解:原式=x2-y2-(2x2-4y2)
=x2-y2-2x2+4y2
=-x2+3y2.
当x=-1,y=eq \f(1,2)时,
原式=-(-1)2+3×(eq \f(1,2))2=-1+eq \f(3,4)=-eq \f(1,4).
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(4)(2019·凉山州)(a+3)2-(a+1)(a-1)-2(2a+4),其中a=-eq \f(1,2);
解:原式=a2+6a+9-(a2-1)-4a-8
=2a+2.
当a=-eq \f(1,2)时,原式=2×(-eq \f(1,2))+2=1.
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(5)[(2m+n)2-(2m-n)(2m+n)-8n]÷2n,其中|2m-1|+(n-3)2=0.
解:原式=(4m2+4mn+n2-4m2+n2-8n)÷2n
=(4mn+2n2-8n)÷2n
=2m+n-4.
因为|2m-1|+(n-3)2=0,
所以2m-1=0,n-3=0.
所以m=eq \f(1,2),n=3.
当m=eq \f(1,2),n=3时,原式=2×eq \f(1,2)+3-4=0.
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类型3 利用整体代入进行化简求值
5.先化简,再求值:(2+a)(2-a)+a(a-5b)+3a5