根号(3-x-)根号(x 1) x²-1当x趋向于1的极限
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![根号(3-x-)根号(x 1) x²-1当x趋向于1的极限](/uploads/image/f/5526904-40-4.jpg?t=%E6%A0%B9%E5%8F%B7%283-x-%29%E6%A0%B9%E5%8F%B7%28x+1%29+x%C2%B2-1%E5%BD%93x%E8%B6%8B%E5%90%91%E4%BA%8E1%E7%9A%84%E6%9E%81%E9%99%90)
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
题是这样的吧:[(√x-√y)^3+2x√x+y√y]/(x√x+y√y)+[3√(xy)-3y]/(x-y)原式=[(x√x-3x√y+3y√x-y√y)+2x√x+y√y]/(x√x+y√y)+[
2/[根号下(-2根3+8)+根3+1]+2/[-根号下(-2根3+8)+根3+1]
原式=2√3+√6-√36+30*√6/6=2√3+√6-6+5√6=2√3+6√6-6
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+根号x+根号(x+2)+根号(x^2+2x)=根号x(根号x+1)+根号(x+2)(根号x+1)=(根号x+1)(根号x+根号(x+2))=3两边同时乘以(根号(x+2)-根号x)得(根号x+1)
根号3X=根号2【x+1】【x-1】3X=2(X^2-1)2X^2-3X-2=0(X-2)(2X+1)=0X1=2X2=-1/2(不合题意,舍去)
(根号3-3根号2)(根号3+根号5)=(根号3-3根号2)*(根号5-根号3)/2=(根号15-3根号10+3根号6-3)/2
原式=[(√x-√y)²+(√x+√y)²]/(√x+√y)(√x-√y)=(x+y-2√xy+x+y+2√xy)/(x-y)=2(x+y)/(x-y)=2(2+√3)/(2-√3
因为x-3>=0,3-x>=0,所以x=3所以√(x-3)+√(3-x)+x^2=√(3-3)+√(3-3)+3^2=9.
√x+1/√x=3,所以√x+1=3√x,1=2√x,所以√x=1/2.则√x-1/√x=(1/2-1)÷1/2=-1(负1)
y=根号(x根号x(根号x))=x^(1/2)*x^(1/4)*x^(1/8)=x^(1/2+1/4+1/8)=x^(7/8)y'=7/8*x^(-1/8)
1/x1+1/x2==(x1+x2)/(x1x2)=(√3+1)/(√3-1)=﹙√3+1)²/2=2+√3
答案为1.直接展开啊
4根号五X根号二=4根号(5X2)=4根号10
(根号2-根号3)x
x^2+3x+1=0x1+x2=-3,x1x2=1,x1
(√5+√3-√2)*(√5-√3+√2)=[√5+(√3-√2)]*[√5-(√3-√2)]=(√5)^2-(√3-√2)^2=5-(3+2-2√6)=2√6