y=-2tan3x+4分之π的单调区间
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=3x²/2x²=3/2
2x-y分之x+y=2,则2x-y=2(x+y)4x-2y分之x+y-4x+4y分之2x-y=2(2x-y)分之x+y-4(x+y)分之2x-y=4(x+y)分之x+y-4(x+y)分之2(x+y)=
lim(x→π/2)tan3x/tanx=lim(x→π/2)3sec^2(3x)/sec^2x=lim(x→π/2)3cos^2x/cosx^2(3x)=lim(y→0)3sin^2y/sin^2(
x+y分之4x-2y-2x-y分之4x+4y=2(2x-y)/(x+y)-4(x+y)/(2x-y)=2*2-4*(1/2)=4-2=2
3/4x=2/3y9x=8yx;y=8:9再问:你几年级呀???再答:呵呵,对的话你就采纳行了。再问:我还是不太明白●▽●再答:3/4x=2/3y9/4x=2y9x=8y再问:完全不懂o_O算了我写啦
周期T=π/w所以周期T=π/3因为周期T=π/3所以得定义域为{x丨x≠π/6+kπ/3k属于Z}值域为R.单调区间不知道,快说你是一班的还是二班的.居然敢上问问来问问题.
(y/4)+2-(2y/6)-3=0方程两边同时乘以12,得3y+24-4y-36=0-y-12=0-y=12y=-12
x+y=x-(x+3/2)^2+21/4=x-x^2-9/4-3x+21/4=-x^2-2x+12/4=-(x+1)^2+4因为-(x+1)^2
1已知函数Y=tan3X的图像的相邻两支截直线Y=pai/3所得的线段长为?解析:∵函数Y=tan3X,∴T=π/3∴所得的线段长为π/32函数Y=sinx的定域义为[a,b],值域为[-1,0.5]
lim(x→π/2)tanx/tan3x=lim(x→π/2)sec^2x/[3sec^2(3x)]=lim(x→π/2)cos^2(3x)/(3cos^2x)=lim(x→π/2)2cos(3x)s
tan3x/2-tanx/2=(sin3x/2)/(cos3x/2)-(sinx/2)/(cosx/2)=(sin3x/2*cosx/2-sinx/2*cos3x/2)/(cos3x/2*cosx/2
sinxisthesameinfinitesimalasx.tan3xisthesameinfinitesimalas3x.Therefore,lim(sinx+tan3x)/2x=4x/2x=2
应用罗必达法则,分子分母同时求导lim(x趋向π)sin2x/tan3x=lim(x趋向π)2cos2x/[3*(sec3x)^2]=2/3*cos2π*(cos3π)^2=2/3*1*(-1)^2=
1、3x=2kπ+x得x=kπ,k∈Z2、x1+x2=-3√3,x1x2=4,由此可知两根为负的.tan(a+b)=(tana+tanb)/(1-tanatanb)=(x1+x2)/(1-x1x2)=
6y-(y+2分之y+2-y分之y)除以y³-4y-2y²+8分之y=6y-[(2y-y²+2y+y²)/(4-y²)]÷y/(y-2)³=
原有的两个解法都不够简洁.直截了当地用tanx=sinx/cosx同角基本关系式不是很容易想到的嘛!原式马上改为:求(sinx/cosx)/(sin3x/cos3x)(先处理正弦,将π/2代入)即简化
tan3x/tanx=(sin3x/sinx)*(cosx/cos3x)lim(cosx/cos3x)=lim-sinx/[(-3)sin3x]=-1/3lim(sin3x/sinx)=-1∴原式=1
证明:tan3x=tan(2x+x)=(tan2x+tanx)/(1-tan2x*tanx)tan3x(1-tan2x*tanx)=tan2x+tanxtan3x-tan3x*tan2x*tanx=t
x/3-y/4=3x+2y==>4x-3y=36x+24y==>①32x=-27y==>x=-27y/32②(4x-3y)^2=(36x+24y)^2②==>16x^2+9y^2-24xy=1296x