初二下册数学两道计算题
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1）2x²y/2mn²除以4xy²/5m²n
=（x²y/4xy²）*（5m²n/2mn²）
=1/4*x/y*(5/2*m/n)
=5mx/8ny
2）x²（x-y）-y²（y-x）+2xy（x-y）
= x²（x-y）+y²（x-y）+2xy（x-y）
=(x-y)(x²+2xy+y²)
=（x-y）(x+y)²
=（x²-y²）（x+y）
2x²y/2mn²除以4xy²/5m²n
=2x²y/2mn²*5m²n/4xy²
=5mx/4yn
x²(x-y)-y²(y-x)+2xy(x-y)
=x²(x-y)+y²(x-y)+2xy(x-y)
=(x-y)(x+y)²
你好 大学理科狗为您解答。1)过程略 只需把前者乘以后面 的倒数即可。约分得5mx/4ny
2）x²（x-y）-y²（y-x）+2xy（x-y）=（x-y)(x²+y²）+2xy（x-y）=（x-y)(x²+y²+2xy)=（x-y)(x+y)²
①解:原式=2xy/2mn×5mn/4xy=5/4 ②解:原式=…………（此题你确定抄完了的？？？？）
①解:原式=2xy/2mn×5mn/4xy=5/4 ②解:原式=…………（此题你确定抄完了的？？？？）
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