已知2x-y=10,求代数式[x²+y²-(x-y)²+2y(x-y)]÷4y的值
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已知2x-y=10,求代数式[x²+y²-(x-y)²+2y(x-y)]÷4y的值
[x²+y²-(x-y)²+2y(x-y)]÷4y
=[x²+y²-[x²-2xy+y²]+2xy-2y²)]÷4y
=[x²+y²-x²+2xy-y²+2xy-2y²]÷4y
=[4xy-2y²]÷4y
=x-y/2=1/2(2x-y)
再通过2x-y=10,有
原式=1/2(2x-y)=1/2×10=5
再问: =[x²+y²-[x²-2xy+y²]+2xy-2y²)]÷4y 括号是怎么样的
再答: =【x²+y²-[x²-2xy+y²]+2xy-2y²】÷4y
再问: 肯定正确吗?
再答: 一定满分啦 如果不相信那就验证 令x=0,y=-10 那么 [x²+y²-(x-y)²+2y(x-y)]÷4y =[0²+(-10)²-(0+10)²+2×(-10)×(0+10)]÷(-40) =(-200)÷(-40) =5 还是对啦 禁得起磨练呀
=[x²+y²-[x²-2xy+y²]+2xy-2y²)]÷4y
=[x²+y²-x²+2xy-y²+2xy-2y²]÷4y
=[4xy-2y²]÷4y
=x-y/2=1/2(2x-y)
再通过2x-y=10,有
原式=1/2(2x-y)=1/2×10=5
再问: =[x²+y²-[x²-2xy+y²]+2xy-2y²)]÷4y 括号是怎么样的
再答: =【x²+y²-[x²-2xy+y²]+2xy-2y²】÷4y
再问: 肯定正确吗?
再答: 一定满分啦 如果不相信那就验证 令x=0,y=-10 那么 [x²+y²-(x-y)²+2y(x-y)]÷4y =[0²+(-10)²-(0+10)²+2×(-10)×(0+10)]÷(-40) =(-200)÷(-40) =5 还是对啦 禁得起磨练呀
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