1.已知x²-10x+y²+4y+29=0,求x²+y²值.2.已知a²+b²-a+4b+(4+1/4)=0,求代数式a²+b²的值.
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![1.已知x²-10x+y²+4y+29=0,求x²+y²值.2.已知a²+b²-a+4b+(4+1/4)=0,求代数式a²+b²的值.](/uploads/image/z/8561358-54-8.jpg?t=1.%E5%B7%B2%E7%9F%A5x%26sup2%3B-10x%2By%26sup2%3B%2B4y%2B29%3D0%2C%E6%B1%82x%26sup2%3B%2By%26sup2%3B%E5%80%BC.2.%E5%B7%B2%E7%9F%A5a%26sup2%3B%2Bb%26sup2%3B-a%2B4b%2B%EF%BC%884%2B1%2F4%EF%BC%89%3D0%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8Fa%26sup2%3B%2Bb%26sup2%3B%E7%9A%84%E5%80%BC.)
1.已知x²-10x+y²+4y+29=0,求x²+y²值.2.已知a²+b²-a+4b+(4+1/4)=0,求代数式a²+b²的值.
1.已知x²-10x+y²+4y+29=0,求x²+y²值.
2.已知a²+b²-a+4b+(4+1/4)=0,求代数式a²+b²的值.
1.已知x²-10x+y²+4y+29=0,求x²+y²值.2.已知a²+b²-a+4b+(4+1/4)=0,求代数式a²+b²的值.
x²-10x+y²+4y+29=x²-10x+25+y²+4y+4=(x-5)^2+(y+2)^2=0
∴x=5 y=-2 ∴x²+y²=25+4=29
2.a²+b²-a+4b+(4+1/4)=a^2-a+1/4+b^2+4b+4=(a-1/2)^2+(b+2)^2=0
∴a=1/2 b=-2 a²+b²=17/4