已知x,y满足x^2+y^2-4x+10y+29=0,求代数式(x+2y)^2-(x-3y)(x+3y)的值
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![已知x,y满足x^2+y^2-4x+10y+29=0,求代数式(x+2y)^2-(x-3y)(x+3y)的值](/uploads/image/z/990177-33-7.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%E6%BB%A1%E8%B6%B3x%5E2%2By%5E2-4x%2B10y%2B29%3D0%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F%28x%2B2y%29%5E2-%28x-3y%29%28x%2B3y%29%E7%9A%84%E5%80%BC)
已知x,y满足x^2+y^2-4x+10y+29=0,求代数式(x+2y)^2-(x-3y)(x+3y)的值
已知x,y满足x^2+y^2-4x+10y+29=0,求代数式(x+2y)^2-(x-3y)(x+3y)的值
已知x,y满足x^2+y^2-4x+10y+29=0,求代数式(x+2y)^2-(x-3y)(x+3y)的值
∵x^2+y^2-4x+10y+29=0,∴(x^2-4x+4)+(y^2+10y+25)=0,
∴(x-2)^2+(y+5)^2=0,∴x=2、y=-5.
∴(x+2y)^2-(x-3y)(x+3y)
=(x+2y)^2-(x^2-9y^2)=(2-10)^2-(4-9×25)=64-4+225=285.
解:
x^2+y^2-4x+10y+29=0
(x-2)^2+(y+5)^2=0
所以
x=2
y=-5
所以
(x+2y)^2-(x-3y)(x+3y)
=64-(x^2-9y^2)
=64-(4-225)
=285