求函数y=x平方 1分之x的四次方加3x的平方 3-搜答案-百度作业网

由题意可得：x+1/x-1=7(x+1/x)^2=64所以x^2+1/x^2=62所以(x^4+x^2+1)/x^2=x^2+1+1/x^2=63再问：我看不懂，您能解释详细点吗？谢谢了再答：（x^2

原式=(x²+y²)/(x²-y²)/[x²(x+y)+y²(x+y)]=(x²+y²)/(x+y)(x-y)/[(x&

展开后分离y=(x^4+x^2+5)/(x^4+x^2+1)=1+4/(x^4+x^2+1)换元令t=x^2则y=1+4/(t^2+t+1)(t∈R)则值域：（0,19/3】再问：我做到了您说的最后一

x+1/x=3两边平方x²+2*x*1/x+1/x²=9x²+2+1/x²=9x²+1/x²=7x²+1/x²=7平方x

x/(x^2-x+1)=71/(x-1+1/x)=7x-1+1/x=1/7x+1/x=8/7x^2/(x^4+x^2+1)=1/(x^2+1+1/x^2)=1/[(x+1/x)^2-1]=1/[(8/

x分之x的平方+x+1=1分之3,分式x的平方分之x的四次方+x的平方+1的值;【x²+x+1】/x=3,即x+1/x=2,(x+1/x)²=4,即x²+1/x²

(1)y'=4*x^3(2)y'=6x/(x^2+5)-(6x^3-4x)/(x^2+5)^2(3)y'=4x+5/2*x^(3/2)(4)y'=4x^3+2x-1/2*x^(-3/2)(5)y'=c

设a=y=(x^4+x²+5)/(x²+1)²,(a-1)x^4+(2a-1)x²+a-5=0,Δ≥0,(2a-1)²-4(a-1)(a-5)≥0,a

d(x²)+d(y^4)+dx+d(2y)=d(1)2xdx+4y³dy+dx+2dy=0所以dy/dx=-(2x+1)/(4y³+2)

y=(x^2+1/2)^2+3/4x^2=0y=1

y'=4x³+6x²+12x+5=04x³+2x²+4x²+12x+5=02x²(2x+1)+(2x+5)(2x+1)=0(2x²

∵x^2+y^2-2x+4y+5=0配方：∴(x-1)^2+(y+2)^2=0那么x-1=0且y+2=0所以x=1,y=-2∴(x^4-y^4)/(2x^2+xy-y^2)*（2x-y)/(xy-y^

学习快乐!

x/(x^2-3x+1)=21/(x-3+1/x)=2x-3+1/x=1/2x+1/x=7/2x^2/(x^4+x^2+1)=1/(x^2+1+1/x^2)=1/[(x+1/x)^2-2+1]=1/[

这是几何画板中y=x^2+1/x的图像,绝对无误,希望能帮助你

x/(x^2-x+1)=61/(x-1+1/x)=6x-1+1/x=1/6x+1/x=1+1/6=7/6(x+1/x)^2=49/36x^2+1/x^2+2=49/36x^2+1/x^2=49/36-

x平方+x+1分之x=二分之一所以x分之（x²＋x＋1）=2x＋1＋x分之1=2x＋x分之1=1平方,得x²＋2＋x²分之1=1x²＋x²分之1=-1

x=7(x²-x+1)7x²-7x+7=x7x²+7=8x两边平方49x^4+98x²+49=64x²两边减去49x²49x^4+49x&s

f(x)=x^2/(1+x^2),将1/x带入得到f(1/x)=1/1+x^2,所以f(x)+f(1/x)=1,(1)1.1(2)1(3)1+1+1+.1=2012

先求导数让导数大于零就是X的单调区间