已知方程组(a²+b²)x²-2b(a+c)x+b²+c²=0,求证b分之c=a分之b=x
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![已知方程组(a²+b²)x²-2b(a+c)x+b²+c²=0,求证b分之c=a分之b=x](/uploads/image/z/3816789-69-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8B%E7%BB%84%28a%26%23178%3B%2Bb%26%23178%3B%29x%26%23178%3B-2b%28a%2Bc%29x%2Bb%26%23178%3B%2Bc%26%23178%3B%3D0%2C%E6%B1%82%E8%AF%81b%E5%88%86%E4%B9%8Bc%3Da%E5%88%86%E4%B9%8Bb%3Dx)
已知方程组(a²+b²)x²-2b(a+c)x+b²+c²=0,求证b分之c=a分之b=x
已知方程组(a²+b²)x²-2b(a+c)x+b²+c²=0,求证b分之c=a分之b=x
已知方程组(a²+b²)x²-2b(a+c)x+b²+c²=0,求证b分之c=a分之b=x
﹙a²+b²﹚x²-2b﹙a+c﹚x+b²+c²=0,
∴a^2x^2-2abx+b^2+b^2x^2-2bcx+c^2=0,
∴(ax-b)^2+(bx-c)^2=0,
a.b.c.x均为实数,
∴ax=b,bx=c.
即有x=b/a=c/b