已知x-7y=0,且y≠0,求2x²﹢xy-3y²分之x²-3xy+y²的值
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![已知x-7y=0,且y≠0,求2x²﹢xy-3y²分之x²-3xy+y²的值](/uploads/image/z/1107706-58-6.jpg?t=%E5%B7%B2%E7%9F%A5x-7y%3D0%2C%E4%B8%94y%E2%89%A00%2C%E6%B1%822x%26%23178%3B%EF%B9%A2xy-3y%26%23178%3B%E5%88%86%E4%B9%8Bx%26%23178%3B-3xy%2By%26%23178%3B%E7%9A%84%E5%80%BC)
已知x-7y=0,且y≠0,求2x²﹢xy-3y²分之x²-3xy+y²的值
已知x-7y=0,且y≠0,求2x²﹢xy-3y²分之x²-3xy+y²的值
已知x-7y=0,且y≠0,求2x²﹢xy-3y²分之x²-3xy+y²的值
答:
x-7y=0
x=7y≠0
所以:x/y=7,y/x=1/7
(x²-3xy+y²)/(2x²﹢xy-3y²)
=(x/y-3+y/x)/(2x/y+1-3y/x)
=(7-3+1/7)/(2*7+1-3/7)
=(29/7)/(102/7)
=29/102