已知实数x,y,z,满足x+y=5,z²=xy+y=9,求x+2y+3Z的值
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已知实数x,y,z,满足x+y=5,z²=xy+y=9,求x+2y+3Z的值
已知实数x,y,z,满足x+y=5,z²=xy+y=9,求x+2y+3Z的值
已知实数x,y,z,满足x+y=5,z²=xy+y=9,求x+2y+3Z的值
x=5-y
z²=(5-y)y+y-9
=6y-y²-9
=-(9-6y+y²)
=-(y-3)²
即z²+(y-3)²=0
要使平方数等于0 则z=0 ; y-3=0
解得
z=0 y=3 x=5-y=2
于是
即原式=2+6+0=8
z²=6y-y²-9=-(y-3)²
z²+(y-3)²=0
∴z=0,y=3
∴x=2
∴原式=8
x+y=5 y=5-x 代入 z^2=x(5-x)+5-x-9 z^2+(x-2)^2=0 则 z=0 x=2 y=3 x+2y+3z=2+2*3=8
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