2017_2018学年八年级数学上册第十四章整式的乘法与因式分解14.1.7单项式除以单项式同步精练(新版)新人教版.docx

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1、1141.7 单项式除以单项式1 aman_ am n_(a0, m, n 都是正整数,且 mn),也就是,同底数幂相除,底数_不变_,指数_相减_2单项式相除,把系数与同底数幂分别_相除_作为商的因式,对于只在被除式里含有的字母,则连同它的指数作为商的一个_因式_3计算:(1)6 x32x2_3 x_;(2)a72a3_0.5 a4_. 易错点睛 已知 x4y5xmyn x2y,求 m n 的值【解】6.【点睛】根据同底数幂的除法法则,则 4 m2,5 n1,m2, n4, m n6.知识点一 a 01(a0)1. 填空:1 0_1_,(10) 0_1_,(2016 )0_1_,2(3.14

2、) 0 ,23 23(a5) 0_1_( a5)2已知( x1) 01,则 x 的取值范围是_ x1_.知识点 二 同底数幂的除法3(2017天门)计算 x3x 的结果是( A )A x2 B2 x2C x4 D34计算:(1) a6a2_ a4_; ( 2)a5a2a_ a2_.5计算:(1)( x2x3)2x6_ x4_; (2)(a2n)2a2_ a4n2 _.知识点三 单项式除以单项式6计算 6a63a2的结果为( D )A3 a4 B3 a3C2 a3 D2 a427计算:(6 a3b4)(3a2b)的结果是( B )A2 B2 ab3 C3 ab3 D2 a5b58若 16a2b4

3、c 除以某个单项式后得 a2b2,这个单项式是( A )A16 b2c B16 bc2C16 ab2 D16 a2b9(2016安徽)计算 a10 a2(a0)的结果是( C )A a5 B a5C a8 D a8 10填空:(1)3 m2n5_6mn2_ mn3;12(2)_5 a5b3c_15a4b ab2c.1311计算:(1)(2016曲靖) a9 a5;【解题过程】解: a4(2)(2016德阳)( a)6a3;【解题过程】解: a3(3)(2016浦东)16 x5y84xy2;【解题过程】解:4 x4y6(4)(2016揭西)5 a3b210a2b;【解题 过程】解: ab12(5

4、)(3 xy)2(x2y);【解题过程】解:9 y(6)( a3b)2(3 a5b2)【解题过程】解: a13312已知 aman a4, aman a2,则 m_3_, n_1_.13若 5x3 y20,则 105x103y_100_.14计算:(1)8a5b4(2 a2b);【解题过程】解: 4 a3b3(2)( 8a3bn 3)(2a2bn);【解题 过程】解:4 ab3(3)(m n)5 ( m n)4;12【解题过程】解:2( m n)(4)(2x2y)3(14x4y3)(7 xy2)【解题过程】解:4 x3y215已知( ambn)3(ab2)2 a4b5,求 m, n 的值(导学号:58024251)【解题过程】解: m2, n3.16(2017邛崃)若( xmx2n)3xm n与 4x2为同类项,且 2m5 n7,求 4m225 n2的值(导学号:58024252)【解题过程】解:原式14.17已知( x1) 2( y2) 2| z3|0.(导学号:58024253)(1)求 x, y, z 的值;(2)求代数式 x2y3z43(xy2z2)26(x2y3z4)2的值【解题过程】解:(1) x1, y2, z3;(2)原式 y1.124

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