求详解.
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求详解.求详解.
求详解.
答:
a和b是二次方程x²-2kx+20=0的两个实数根
根据韦达定理有:
a+b=2k
ab=20
所以:
(a+1)²+(b+1)²
=a²+2a+1+b²+2b+1
=(a+b)²-2ab+2(a+b)+2
=(2k)²-2×20+2×2k+2
=4k²+4k-38
=4(k²+k+1/4)-39
=4(k+1/2)²-39
当且仅当k+1/2=0即k=-1/2时取得最小值-39
a+b=2k,ab=20用得伟达定理又∵a²+2a+1+b²+2b+1=²-2ab+2+2∴4k²-40+4k+2=4k²+4k-38,再用二次函数对称轴有最小值k=-1/2