已知方程x²+根号下2*x-1=0有2个根x1、x2 求①x1²+x2²还有②(x1-x2)²③x1负一次方+x2的负一次方希望解答一下,必有重谢
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![已知方程x²+根号下2*x-1=0有2个根x1、x2 求①x1²+x2²还有②(x1-x2)²③x1负一次方+x2的负一次方希望解答一下,必有重谢](/uploads/image/z/11472618-66-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8Bx%26%23178%3B%2B%E6%A0%B9%E5%8F%B7%E4%B8%8B2%2Ax-1%3D0%E6%9C%892%E4%B8%AA%E6%A0%B9x1%E3%80%81x2+%E6%B1%82%E2%91%A0x1%26%23178%3B%2Bx2%26%23178%3B%E8%BF%98%E6%9C%89%E2%91%A1%EF%BC%88x1-x2%EF%BC%89%26%23178%3B%E2%91%A2x1%E8%B4%9F%E4%B8%80%E6%AC%A1%E6%96%B9%2Bx2%E7%9A%84%E8%B4%9F%E4%B8%80%E6%AC%A1%E6%96%B9%E5%B8%8C%E6%9C%9B%E8%A7%A3%E7%AD%94%E4%B8%80%E4%B8%8B%2C%E5%BF%85%E6%9C%89%E9%87%8D%E8%B0%A2)
已知方程x²+根号下2*x-1=0有2个根x1、x2 求①x1²+x2²还有②(x1-x2)²③x1负一次方+x2的负一次方希望解答一下,必有重谢
已知方程x²+根号下2*x-1=0有2个根x1、x2 求①x1²+x2²
还有②(x1-x2)²③x1负一次方+x2的负一次方希望解答一下,必有重谢
已知方程x²+根号下2*x-1=0有2个根x1、x2 求①x1²+x2²还有②(x1-x2)²③x1负一次方+x2的负一次方希望解答一下,必有重谢
由韦达定理得
x1+x2= -根号下2
x1*x2 =-1
①x1²+x2²=x1²+x2²+2x1*x2-2x1*x2
=(x1+x2)²-2x1*x2=2-(-1)=4
②(x1-x2)²=x1²+x2²-2x1*x2=4-2(-1)=6
③x1负一次方+x2的负一次方=(x1+x2)/x1*x2
=根号下2
x²+根号下2x-1=0
x1+x2=-根号下2
x1x2=-1
x1²+x2²=(x1+x2)²-2x1x2=2+2=4
(x1-x2)²=x1²+x2²-2x1x2=4+2=6
1/x1+1/x2=(x1+x2)/x1x2=根号下2
①x1²+x2²=(x1+x2)²-2x1x2 =(—根号下2)²-2乘以(-1)=4
②(x1-x2)²=(x1+x2)²-4x1x2 =(—根号下2)²-4乘以(-1)=6
③x1负一次方+x2的负一次方=(x1x2)分之(x1+x2)=2分之根号下2
方程x²+根号下2*x-1=0有2个根x1、x2
∴x1x2=-1
x1+x2=-√2
①x1²+x2² =(x1+x2)²-2x1x2=2-(-2)=4
②(x1-x2)²=(x1+x2)²-4x1x2=2-(-4)=-6
③x1负一次方+x2的负一次方=(x1+x2)/x1x2=√2