设x1、x2是方程4²-7x-3=0的两根,不解方程,求下列各式的值(要过程)(1)(x1+2)(x2+2) (2)x1²× x2 +x1× x2²
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![设x1、x2是方程4²-7x-3=0的两根,不解方程,求下列各式的值(要过程)(1)(x1+2)(x2+2) (2)x1²× x2 +x1× x2²](/uploads/image/z/3762923-59-3.jpg?t=%E8%AE%BEx1%E3%80%81x2%E6%98%AF%E6%96%B9%E7%A8%8B4%26%23178%3B-7x-3%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%2C%E4%B8%8D%E8%A7%A3%E6%96%B9%E7%A8%8B%2C%E6%B1%82%E4%B8%8B%E5%88%97%E5%90%84%E5%BC%8F%E7%9A%84%E5%80%BC%EF%BC%88%E8%A6%81%E8%BF%87%E7%A8%8B%EF%BC%89%EF%BC%881%EF%BC%89%EF%BC%88x1%2B2%EF%BC%89%EF%BC%88x2%2B2%EF%BC%89+%EF%BC%882%EF%BC%89x1%26%23178%3B%C3%97+x2+%2Bx1%C3%97+x2%26%23178%3B)
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设x1、x2是方程4²-7x-3=0的两根,不解方程,求下列各式的值(要过程)(1)(x1+2)(x2+2) (2)x1²× x2 +x1× x2²
设x1、x2是方程4²-7x-3=0的两根,不解方程,求下列各式的值(要过程)
(1)(x1+2)(x2+2) (2)x1²× x2 +x1× x2²
设x1、x2是方程4²-7x-3=0的两根,不解方程,求下列各式的值(要过程)(1)(x1+2)(x2+2) (2)x1²× x2 +x1× x2²
因为x1,x2是方程4x²-7x-3=0
所以x1+x2=7/4
x1x2=-3/4
(x1+2)(x2+2)=x1+x2+x1x2+4=7/4-3/4+4=5
x1²× x2 +x1× x2²=x1x2(x1+x2)=7/4(-3/4)=-21/16
如有不明白,
....楼主你的题目好像打错了
x1、x2是方程4²-7x-3=0的两根,则x1+x2 = -b/a = 7/4; x1x2 = c/a = -3/4;
(1)(x1+2)(x2+2) = x1x2+2(x1+x2)+4 = -3/4 + 2*7/4+4 = 4又11/4;
(2)x1²× x2 +x1× x2² = x1x2(x1+x2) = -3/4*7/4 = -21/16;
x1+x2=7/4, x1*x2=-3/4.
(1)(x1+2)(x2+2)=x1*x2+2(x1+x2)+4=-3/4+2*7/4+4=7.75.
(2)x1²× x2 +x1× x2²=x1*x2(x1+x2)=-3/4*7/4=-21/16.