解方程x2 x-6=0,x2-根号3x-4分之1=0
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/02 07:29:28
移项,得x2-6x=2配方,得(x-3)2=11,即x-3=11或x-3=-11,所以,方程的解为x1=3+11,x2=3-11.
方程的两边同乘(2x-3)(2x+3),得2x(2x+3)-(2x-3)=(2x-3)(2x+3),解得x=-3.检验:把x=-3代入(2x-3)(2x+3)=27≠0.∴原方程的解为:x=-3.
2/(x2-x)+6/(1-x2)=7/(x2+x)2/x(x-1)-6/(x-1)(x+1)=7/x(x+1)[x*(x-1)*(x+1)]*[2/x(x-1)-6/(x-1)(x+1)]=[7/x
△=(-6)2-4×4×(-3)=60,x=6±2212×4=3±214,所以x1=3+214,x2=3−214.
2x2x+5−55x−2=1.2x(5x−2)(2x+5)(5x−2)−5(2x+5)(2x+5)(5x−2)=1,2x(5x−2)−5(2x+5)(2x+5)(5x−2)=1,10x2-14x-25
原式可化为1/(x+1)(x-3)+2/(x-3)((x+2)+3/(x+1)(x+2)=0两边同乘以:(x+1)(x+2)(x-3)得:(x+2)+2(x+1)+3(x-3)=0(x≠-1,x≠-2
x------1x------5(x+1)(x+5)=0x1=-1x2=-5
等式两边同时乘以(x+3)(x-2)(x+2)就可以去分母了
7/(X2+X)+3/(X2-X)=6/(X2-X),去分母,等式两端同时乘X(X+1)(X-1):7(X-1)+3(X+1)=6(X+1),7X-7+3X+3=6X+6,7X+3X-6X=6+7-3
韦达定理啊!x1+x2=-5,x1*x2=6再换算即可
2x+5x+3+4-x/2+2x+x-2=019x/2=-5x=-19/10检验:带入x是、原式=0
(x²-2x)+(x²-2x)-2=0(x²-2x+2)(x²-2x-1)=0∵x²-2x+2=(x-1)²+1>0∴x²-2x-
设0
化为(x+2)(x-3)=0x=-2或x=3所以解集为{-2,3}
7/(x+x2)-3/(x-x2)=6/(x2-1)两边同乘以x(x+1)(x-1),得7(x-1)+3(x+1)=6x7x-7+3x+3=6x10x-6x=3-74x=-4x=-1经检验x=-1是增
令x²+x=t原方程变为t+1=6/tt²+t-6=0(t+3)(t-2)=0则t=2或-31)x²+x=2x²+x-2=0(x+2)(x-1)=0x=-2或x
x2-5x-6=0,∴(x-6)(x+1)=0,∴x-6=0或x+1=0,∴x1=6,x2=-1.
X1X2X3X4X5X6X7sum50681181863044907942010这是我通过EXCEL算出来的结果X1+X2+X3=236
答:x^3-x^2-5x+6=0(x^3-2x^2)+(x^2-5x+6)=0(x-2)x^2+(x-2)(x-3)=0(x-2)(x^2+x-3)=0所以:x-2=0或者:x^2+x-3=0解得:x
-x^2+4x=0-x(x-4)=0-x=0x-4=0x=0或者x=4