已知(m^2+n^2)(m^2+n^2-9)-10=0,求代数式m^2+n^2的值设M²+N²=x(x>0),则原方程变为x(x-9)-10=0 解得x=10(x=-1舍去)所以M²+N²=10
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![已知(m^2+n^2)(m^2+n^2-9)-10=0,求代数式m^2+n^2的值设M²+N²=x(x>0),则原方程变为x(x-9)-10=0 解得x=10(x=-1舍去)所以M²+N²=10](/uploads/image/z/2091244-4-4.jpg?t=%E5%B7%B2%E7%9F%A5%28m%5E2%2Bn%5E2%29%28m%5E2%2Bn%5E2-9%29-10%3D0%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8Fm%5E2%2Bn%5E2%E7%9A%84%E5%80%BC%E8%AE%BEM%26%23178%3B%2BN%26%23178%3B%3Dx%EF%BC%88x%EF%BC%9E0%EF%BC%89%2C%E5%88%99%E5%8E%9F%E6%96%B9%E7%A8%8B%E5%8F%98%E4%B8%BAx%EF%BC%88x-9%EF%BC%89-10%3D0%E3%80%80%E3%80%80%E8%A7%A3%E5%BE%97x%3D10%EF%BC%88x%3D-1%E8%88%8D%E5%8E%BB%EF%BC%89%E6%89%80%E4%BB%A5M%26%23178%3B%2BN%26%23178%3B%3D10)
已知(m^2+n^2)(m^2+n^2-9)-10=0,求代数式m^2+n^2的值设M²+N²=x(x>0),则原方程变为x(x-9)-10=0 解得x=10(x=-1舍去)所以M²+N²=10
已知(m^2+n^2)(m^2+n^2-9)-10=0,求代数式m^2+n^2的值
设M²+N²=x(x>0),则原方程变为x(x-9)-10=0 解得x=10(x=-1舍去)所以M²+N²=10
已知(m^2+n^2)(m^2+n^2-9)-10=0,求代数式m^2+n^2的值设M²+N²=x(x>0),则原方程变为x(x-9)-10=0 解得x=10(x=-1舍去)所以M²+N²=10
是对的.非常正确.
这种方法叫作“换元法”
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