已知x,y互为相反数,且(2x-3)^2-(2y+1)^2求x,y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
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![已知x,y互为相反数,且(2x-3)^2-(2y+1)^2求x,y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](/uploads/image/z/578253-21-3.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%2C%E4%B8%94%282x-3%29%5E2-%282y%2B1%29%5E2%E6%B1%82x%2Cy%E7%9A%84%E5%80%BC%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)
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已知x,y互为相反数,且(2x-3)^2-(2y+1)^2求x,y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互为相反数,且(2x-3)^2-(2y+1)^2求x,y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互为相反数,且(2x-3)^2-(2y+1)^2求x,y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
由条件可得:x²--4x+4+y²+10y+25=0得:(x--2)²+(y+5)²=0,所以x=2,y=--5代人化简后的代数式可得结果:285
第一题,题目不明
第二题,x=2,y=-5,答案为365
(1)不会
(2)把y=-x带入要求的代数式就可以了,x=-1,y=1
(3)...??原来是第一题