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已知方程3x²+4x-5=0的两根为X1.X2,求下列各式的值;
(1)X1²+X2² (2)1/X1+1/X2 (3)(X1-X2)² (4)(X1+1)(X2+1)
人气:111 ℃ 时间:2020-04-18 03:31:52
解答
据根与系数的关系有:x1+x2=-4/3 x1x2=-5/3 (1)X1²+X2²=(x1+x2)^2-2x1x2=16/9+10/3=46/9(2)1/X1+1/X2 =(x2+x1)/x1x2=(-4/3)/(-5/3)=4/5(3) (x1-x2)^2=(x1+x2)^2-4x1x2=16/9+20/3=76/9(4) (X1+...
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