作业帮已知x1,x2是方程x²+4[kx+(1-4k)]²=4的两根,求(x1-x2)²
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已知x1,x2是方程x²+4[kx+(1-4k)]²=4的两根,求(x1-x2)²
已知x1,x2是方程x²+4[kx+(1-4k)]²=4的两根,求(x1-x2)²
已知x1,x2是方程x²+4[kx+(1-4k)]²=4的两根,求(x1-x2)²
x²+4[kx+(1-4k)]²=4
x²+4[k²x²+2(1-4k)kx+(1-4k)²]=4
(4k²+1)x²-8(4k-1)kx+4(1-4k)²-4=0
(4k²+1)x²-8(4k-1)kx+16k²-8k=0
x1+x2=8(4k-1)k/(4k²+1) x1*x2=(16k²-8k)/(4k²+1)
(x1-x2)²
=(x1+x2)²-4x1*x2
=[8(4k-1)k/(4k²+1) ]²-4(16k²-8k)/(4k²+1)
化简即可
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