初二数学题,有过程1题 5/12(x+y)^2*(-x-y)^3*[-3/4(x-y)^3]*6/5(y-x)^2 2题 [(x-y)^3]^6*(y-x)^3*3(x-y)^2
2019-05-29
初二数学题,有过程
1题 5/12(x+y)^2*(-x-y)^3*[-3/4(x-y)^3]*6/5(y-x)^2
2题 [(x-y)^3]^6*(y-x)^3*3(x-y)^2
优质解答
1题
5/12(x+y)^2*(-x-y)^3*[-3/4(x-y)^3]*6/5(y-x)^2
= -5/12(x+y)^2*(x+y)^3*[-3/4(x-y)^3]*6/5(x-y)^2
=(-5/12)*(-3/4)*(6/5)(x+y)^5(x-y)^5
=(3/8)(x+y)^5(x-y)^5
2题
[(x-y)^3]^6*(y-x)^3*3(x-y)^2
=(x-y)^18*[-(x-y)^3]*3(x-y)^2
=-3(x-y)^23
提示:同底数幂相乘,底数不变,指数相加
1题
5/12(x+y)^2*(-x-y)^3*[-3/4(x-y)^3]*6/5(y-x)^2
= -5/12(x+y)^2*(x+y)^3*[-3/4(x-y)^3]*6/5(x-y)^2
=(-5/12)*(-3/4)*(6/5)(x+y)^5(x-y)^5
=(3/8)(x+y)^5(x-y)^5
2题
[(x-y)^3]^6*(y-x)^3*3(x-y)^2
=(x-y)^18*[-(x-y)^3]*3(x-y)^2
=-3(x-y)^23
提示:同底数幂相乘,底数不变,指数相加