arctanx x2(1 x2)

要使根号(x2+2x+4)-根号(x2-x+1)

设(x²-1)/(x²+2x)=t则8t+3/t=118t²-11t+3=0(8t-3)(t-1)=0解得t=3/8或t=11.t=3/8(x²-1)/(x&#

去分母得：x^2(y-1)+x(1-y)+y=0y=1时,上式无解y=1时,为二次式,须有delta>=0即(1-y)^2-4y(y-1)>=0(y-1)(3y+1)再问：x^2(y-1)+x(1-y

x2+3x2+1=0中的3x2表示什么?再问：已知X2+3X+1=0，求X2+1/X2的值？得数是7。求过程？我打的是X的平方。怎么会出X2、再答：答：因为x≠0,两边都除以x得：x+1/x=-3，两

由于1=x2+y2+z2=（x2+12y2）+（12y2+z2）≥2x•y2+2•y2•z=2（xy+yz），当且仅当x=y2=z时，等号成立，∴x=y2=z=12时，xy+yz的最大值为22．故答案

∫arctanxdx/[x^2(1+x^2)]=∫arctanxdx/x^2-∫arctanxdx/(1+x^2)=∫arctanxd(-1/x)-∫arctanxdarctanx=-(arctanx

3/x2=1/x2-x即3*2-3x=x*22x*2=3Xx=0（舍去）x=3/2

解题思路：吸纳化简，根据已知条件，整体代入可解。解题过程：

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

令a=x2+x(a+1)(a+12)=42a2+13a+12=42a2+13a-30=0(a+15)(a-2)=0a=-15,a=2x2+x=-15x2+x+15=0无解x2+x=2x2+x-2=(x

(X^2-y+1)(X^2+1)+X^2y+y-X^2=(X^2-y+1)(X^2+1)+(X^2+1)y-X^2=(X^2-y+y+1)(X^2+1)-X^2=(X^2+1)^2-x^2=(x^2+

y=x2\x2+1=[(x^2+1)-1]/(x^2+1)=1-1/(x^2+1)x^2+1>=11/(x^2+1)属于(0,1]所以原函数值域为[0,1)

（x2+3/根号x2+1）^2-(2根号2)^2=(x^4-2x^2+1)/8(x^2+1)=(x^2-1)/8(x^2+1)>=0,又因为不等式两边均为正,所以x2+3/根号x2+1≥2根号2

(x^2+1)^2-4x(x^2-1)=(x^2-1)^2-4x(x^2-1)+4x^2=[x^2-1-2x]^2

x趋近无穷?如果是无穷,答案是1/2先有理化,然后再分子分母各除以x

f(x)=-(x²+1-2)/(x²+1)=-(x²+1)/(x²+1)+2/(x²+1)=-1+2/(x²+1)x²>=0x&s

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为x的平方,y=(x2-1)/(x2+1)两边同乘以x2+1得：y(x2+1)=x2-1去括号y*x2+y=x2-1移项y*x2-x2+y+1=0(y-1)x2+y+1=0x为实数,x的方程有实数

Xn/（x1+x2+...Xn-1)(X1+X2...+Xn）=1/(x1+x2+...+xn-1)-1/(x1+x2+...+xn-1+xn)所以原式=1/x1-1/(x1+x2)+1/(x1+x2

令a=x+1/xa²=x²+2+1/x²2(a²-2)-9a+14=0(2a-5)(a-2)=0x+1/x=5/22x²-5x+2=0(2x-1)(x