设X1X2是方程2x²+4x-3=0的两根.求(x1+2)*(x2+2) x2/x1+x1/x2
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/20 02:31:49
![设X1X2是方程2x²+4x-3=0的两根.求(x1+2)*(x2+2) x2/x1+x1/x2](/uploads/image/z/5248281-57-1.jpg?t=%E8%AE%BEX1X2%E6%98%AF%E6%96%B9%E7%A8%8B2x%26%23178%3B%2B4x-3%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9.%E6%B1%82%EF%BC%88x1%2B2%EF%BC%89%2A%EF%BC%88x2%2B2%EF%BC%89+x2%2Fx1%2Bx1%2Fx2)
x͒JA_`n
D !uGXT ~A ~DP~̜ݽ.`.Μ93H瞱G>vZ_3wvqK3[mGl9OlˈD`]d-CHD_(WTX?xo$A#"TkD _(A|)C}Μ)■'ET=pJp",WA8 &^K
P5
Le eSF1ofM[~w'H7}z6h%,J̜X>EF{Xk1;3KO|c>sZ1><ǻ'>+}bb F|x%`y߽n1ov-
>X;eB~FcJɡ~>
设X1X2是方程2x²+4x-3=0的两根.求(x1+2)*(x2+2) x2/x1+x1/x2
设X1X2是方程2x²+4x-3=0的两根.
求(x1+2)*(x2+2) x2/x1+x1/x2
设X1X2是方程2x²+4x-3=0的两根.求(x1+2)*(x2+2) x2/x1+x1/x2
设X1X2是方程2x²+4x-3=0的两根.
满足 x1+x2 = -2 x1x2 = -3/2
求(x1+2)*(x2+2)
= x1x1 + 2 (x1+x2) +4
= -3/2 - 4 +4
= -2分之3
x2/x1+x1/x2
= (x1²+x2²) /x1x2
=[ (x1+x2) ² - 2x1x2 ] /x1x2
= ( 4 + 3 ) / (-3/2 )
= 7 x ( -2/3)
= - 3分之 14
施主,我看你骨骼清奇,
器宇轩昂,且有慧根,
乃是万中无一的武林奇才.
潜心修习,将来必成大器,
鄙人有个小小的考验请点击在下答案旁的
"选为满意答案"