x2+7x+12=0

解一元二次方程,谢x2+12x=0,4x2-1=0,x2=7x,x2-4x-21=0,（x-1）（x+3）=12,3x2+2x-1=0x2+12x=0,x(x+12)=0x=0,-124x2-1=0,(2x+1)(2x-1)=0x=-1/2

解方程:(3x2-2x+1)(3x2-2x-7)+12=0(3x2-2x+1)(3x2-2x-7)+12=0用十字相乘法设a=3x²-2x-3将a代入原方程(3x²-2x+1)(3x²-2x-7)+12=0,得

(x2-x)2-8(x2-x)+12=0                &nbs

7/x2+x+1/x2-x=0解x吗?原式：=7/2x+1/2x=0即8/2x=0解得x=4

解一元二次方程,196x2-1=0,4x2+12x+9=81,x2-7x-1=0,2x2+3x=3,x2-2x+1=25,x（2x-5）=4x-10①∵196x²-1=0∴x²＝1/196∴X＝±1/14②∵4x

7/x2+x+1/x2-x=6/x2-17/x(x+1)+1/x(x-1)=6/(x+1)(x-1)两边乘x(x+1)(x-1)7x-7+x+1=6x2x=6x=3经检验,x=3是方程的解对分母因式分7/x(x+1)+1/x(x-1)=6/

x1x2是方程3x2-7x+2=0两根求x1+x23x²-7x+2=0)3x-1)(x-2)=0x1=1/3,x2=2所以x1+x2=7/3x1x2是方程3x²-7x+2=0两根那么由韦达定理有x1+x2=7/3如果不懂

x1x2是方程3x2-7x+2=0两根求x1-x2（3x-1）(x-2)=0x1=2x2=1/3x1-x2=5/33x^2-7x+2=0(3x-1)(x-2)=0x1=1/3x2=2x1x2是方程3x2-7x+2=(3x-1)(x-2)=0

x2-3x+1=0x2/(x4-x2+1)请问这是什么啊?希望能把问题完善了好吧?

x2-5x+1=0则x2+x2/1你可以参见“韦达定理”方程两个根的积是1,说明他们互为倒数.x^2+1/x^2=(x+1/x)^2-2*x*1/x=（-5）²-2=23

(1)3x2-x-40x2-x-12=0x2+3x-4016-8x+x2=0(2)解关于x的不等式x2+2x+1-a2=0(a为常数)(3x-4)(x+1)0,x＜-1或x4/3,(x-4)(x+3)＜=0,-3＜=x＜=4,(x+4)(x

x2-3x-1=0,求①x21/x2;②x2-1/x2∵x2-3x-1=0∴X-2-1/X=0∴X-1/X=3∴①平方,X²-2+1/X²=9∴X²+1/X²=11②∵(X+1/X)²=X&

7x的二次方-6x-2=0的两根x1.x2.求|x1-x2|=x1-x2分之（x1+x2)=x2分之x1=每一个等号后面是一个问题由题得,x1+x2=6/7,x1*x2=-2/7所以|x1-x2|=√[(x1+x2)^2-4x1*x2]=2

4X2-4X+1=0X2-12X+35=0X2-2X-35=01,原式=（2x-1）²=0所以x1=x2=0.52,原式=（x-7）（x-5）=0所以x1=7x2=53,原式=（x-7）（x+5）=0所以x1=7x2=-5关键问题

解方程4x2-4x+1=x2+6x+97x2-√6x-5=0 1楼

4x2-12x-7=0因式分解4x^2-12x-7=0(2x+1)(2x-7)=0x=-1/2或x=7/2

已知集合M={x|x2-x-6≤0},N={x|x2-7x+12≥0},则"x∈M"是"x∈N"的A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件M{-2≤x≤3},N{x≥4||x≤3},所以"x∈M"-->"x∈

7x2+44x+1=0x=-15/44求X?X=-(15/44)

x2+4x-7=0x²+4x-7=0(x²+4x+4)=11(x+2)²=11x+2=±√11解得x=-2±√11

一元二次方程的解法（要有过程）⑴x2-8x+12+0⑵x2-6x=7⑶x2-2x=7⑷x2-x=2⑸x2+3x=3x2是x的平方x²-8x+12=0(x-6)(x-2)=0x=6或x=2x²-6x=7x²-6x