已知x1=2分之-1 根号5,x2=2分之-1-根号5
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/18 22:37:11
![已知x1=2分之-1 根号5,x2=2分之-1-根号5](/uploads/image/f/4227260-68-0.jpg?t=%E5%B7%B2%E7%9F%A5x1%3D2%E5%88%86%E4%B9%8B-1+%E6%A0%B9%E5%8F%B75%2Cx2%3D2%E5%88%86%E4%B9%8B-1-%E6%A0%B9%E5%8F%B75)
有题意有:x1x2=m,x1+x2=2根号2,2x1+x2=-3根号2+1解得x1=-5根号2+1,x2=7根号2-1,m=-71+12根号2;(根号下x1/x2)+(根号下x2/x1)无解,因为x1
a,b是方程x^2+5x+2=0的两个根,则由韦达定理有:a+b=-5,ab=2设S=根号(b/a)+根号(a/b)S^2=b/a+a/b+2=[(a+b)^2-2ab]/ab+2=[(-5)^2-4
根号内必须大于等于0故有x-1≥0且1-x≥0即x≥1且x≤1所以x=1将x=1代回去得y=3然后将x,y代入所求式即可你的所求式表述不是很清楚,所以没办法帮你求了
x=(√7+√5)/2,y=(√7-√5)/2,xy=1/2,x+y=√7y/x+x/y=(x^2+y^2)/xy=[(x+y)^2-2xy]/xy=(7-1)/(1/2)=12
x1+x2=5;x1x2=1;(1)x1/x2+x2/x1=(x1²+x2²)/(x1x2)=((x1+x2)²-2x1x2)/(x1x2)=(25-2)/1=23;(2
已知一元二次方程x2-(根号3+1)x+根号3-1=0的两根为x1,x2则由韦达定理x1+x2=√3+1x1*x2=√3-1所以1/x1+1/x2=(x1+x2)/(x1*x2)=(√3+1)/(√3
(√x-√y)²+4√(xy)=(√x+√y)²x=2/5y=8/5√(xy)=4/5√x+√y=√(2/5)+2√(2/5)=3√(2/5)(√x+√y)²/[x+√(
第二问后面5x是x1还是x2再问:我再写一遍吧(1)求x1/x2+x2/x1;(2)求x1^2+5X2,是x2再答:
x分之y+y分之x=(x²+y²)/xy=[(x+y)²-2xy]/xy=(x+y)²/xy-2=[¼(√7+√5+√7-√5)²]/[
因为x=2分之1(根号5+根号3),y=(根号5-根号3),所以x²=2+根号15y²=2-根号15xy=1/2所以x方-xy+y方=4-1/2=3又1/2y分之x+x分之y=xy
韦达定理得:X1+X2=根号3+1X1X2=根号3-11/X1+1/X2=(X1+X2)/(X1X2)=(根号3+1)/(根号3-1)=(根号3+1)(根号3+1)/[(根号3-1)(根号3+1)]=
y=x/2-√(x-2)+2√(2-x)被开方数大于等于0x-2≥0且2-x≥0x≥2且x≤2所以x=2y=1√(4x+y)-√(x²+5y)=√(4*2+1)-√(2²+5)=0
根号X-根号X分之一=2二边平方得:x-2根号X*根号1/X+1/x=4即X+1/X=4+2=6平方得:X^2+2+1/X^2=36所以,X^2+1/X^2=34
x=2分之根号3-根号2y=2分之根号3+根号2xy=1/4y-x=根号2x分之1-y分之1=(y-x)/xy=根号2/(1/4)=4根号2
由韦达定理得x1+x2=-根号下2x1*x2=-1①x1²+x2²=x1²+x2²+2x1*x2-2x1*x2=(x1+x2)²-2x1*x2=2-(
2√2-2或-2√2-2
x1+x2=-3/2x1x2=-21/x1+1/x2=(x1+x2)/x1x2=(-3/2)/(-2)=3/4x1²+x2²=(x1+x2)²-2x1x2=(-3/2)&
x1+x2=4x1x2=-1(x1+x2)^2/(1/x1+1/x2)=(x1+x2)^2*x1x2/(x1+x2)=x1x2*(x1+x2)=-4