计算:(1)6a5b6c4÷(-3a2b3c)÷(2a3b3c3).(2)(x-4y)(2x+3y)-(x+2y)(x-y).
题型:解答题难度:一般来源:不详
计算:(1)6a5b6c4÷(-3a2b3c)÷(2a3b3c3). (2)(x-4y)(2x+3y)-(x+2y)(x-y). |
答案
(1)6a5b6c4÷(-3a2b3c)÷(2a3b3c3), =-2a2b3c÷2a3b3c3, =-1;
(2)(x-4y)(2x+3y)-(x+2y)(x-y), =2x2+3xy-8xy-12y2-(x2-xy+2xy-2y2), =2x2-5xy-12y2-x2-xy+2y2, =x2-6xy-10y2. |
举一反三
若(3m-2)x2yn+1是关于x,y的系数为1的五次单项式,则m=______,n=______. |
单项式-y3x2z与24x5y的积为( )A.-4x7y4z | B.-4x7y4 | C.-3x7y4z | D.3x7y4z |
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由102×(2×103)=2×102×103=2×105这样的式子不难想到,x2(2x3)=2x5 (1)阅读并在每条横线上写出得出该式的依据. (6an-1)(-2ab) =6an-1(-2)a•b ①______, =-12(an-1a)•b ②______, =-12an-1+1•b ③______, =-12anb (2)仿照上面解题过程求a2b2与ab3c5的乘积. |
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