已知x+y=1求x的三次方+y的三次方+3xy的值
解:
x+y=1
x³+y³+3xy
=(x+y)(x²-xy+y²)+3xy(立方和公式)
=x²-xy+y²+3xy(将x+y=1代入)
=x²+2xy+y²
=(x+y)²
=1²
=1
若:x+y=1
则:x^3+y^3=(x+y)(x^2-xy+y^2)=x^2-xy+y^2
x^3+y^3+3xy=x^2-xy+y^2+3xy=x^2+2xy+y^2=(x+y)^2=1
∵(x + y)³=x³+3x²y+3xy²+y³
∴x³+y³=(x + y)³-(3x²y+3xy²)
∴x³+y³+3xy
=(x + y)³+3xy-3x²y-3xy²
=(x + y)³+3xy(1-x-y)
=(x + y)³+3xy(1-(x+y))
=1+3xy(1-1)
=1
显然条件不足,你可以分别以0,1带入,以1/2,1/2代入,两种结果不同,所以这个题目显然条件不足
x³+y³+3xy
=(x+y)(x²-xy+y²)+3xy
=1×((x²-xy+y²)+3xy
=x²+2xy+y²
=(x+y)²
=1
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