已知x²-4x+y²-10y+29=0求x²+2x³y²+x⁴y²的值
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已知x²-4x+y²-10y+29=0求x²+2x³y²+x⁴y²的值
已知x²-4x+y²-10y+29=0求x²+2x³y²+x⁴y²的值
已知x²-4x+y²-10y+29=0求x²+2x³y²+x⁴y²的值
x²-4x+y²-10y+29=0
x²-4x+4+y²-10y+25=0
(x-2)²+(y-5)²=0
x-2=0 y-5=0
x=2 y=5
于是
x²+2x³y²+x⁴y²
=x²(1+2xy²+x²y²)
=2²×(1+2×2×5²+2²×5²)
=4×(1+100+100)
=804
x²-4x+y²-10y+29=0
x²-4x+4+y²-10y+25=0
(x-2)²+(y-5)²=0
(x-2)²=0,(y-5)²=0
x=2
y=5
x²+2x³y²+x⁴y²
=x²(1+2xy²+x²y²)
=2²*(1+2*2*5²+2²*5²)
=4*(1+100+100)
=4*201
=804
x^2-4x+16+y^2-10y+25=0
(x-2)^2+(y-5)^2=0
x=2 y=5
代入原式即解得