y=sin根号下x 根号下1-新的平方-搜答案-搜搜考试网

√xy－√x/y＝√5·4－√5/4＝2√5－√5/2＝3√5/2

两式相减得：x-y+√2（√y-√x）=0即x-y=√2（√x-√y）因为x不等于y,所以√x-√y≠0所以,√x+√y=√2两式相加得：x+y+√2（√x+√y）=2√3,所以x+y=2√3-2又x

1.〔2kπ,π／2＋2kπ）2.〔－tan1,tan1〕

由题意可得sin(2x+π/6)

[(2/3)x√(9x)+6x√(y/x)]+[y√(x/y)-x²√(1/x)]化简：原式=[(2/3)*3*x√x+6√(xy)]+[√(xy)-x√x]=2x√x+6√(xy)+√(x

设z=根号下1+x平方,则dy/dx=(dy/dz)*(dz/dx)=(cos根号下1+x的二次方)*x/(1+x二次方)

y=xarcsin√[x/(1+x)]+arctan√(x-√2)-√x,求导dy/dx=arcsin√[x/(1+x)]+x{√[x/(1+x)]}′/√[1-x/(1+x)]+[√(x-√2)]′

d/dx[sin^-1(x)+x√(1-x²)]=d/dxsin^-1(x)+d/dxx√(1-x²)=1/√(1-x²)+[√(1-x²)+x*(-x)/√(

=-1-4³√x-4³√y=-1-4(³√x+³√y)=-1-4*(-7)=27

(1)y=√(2sinx+1)2sinx+1>=0sinx>=-1/2所以x取值范围为[-π/6+2kπ,7π/6+2kπ](k为整数)(2)y=√（16-x^2）+√（-cosx）16-x^2>=0

结果为根号下x+根号下y解2xy/(x根号下y+y根号下x）分母提公因式根号下xy然后前后两式分母都含根号下x+根号下y合并后约分得根号下x+根号下y

√[(sinx)^4+4(cosx)^2]-√[(cosx)^4+4(sinx)^2]=√[((sinx)^2-2)^2]-√[((cosx)^2-2)^2]=(sinx)^2-2-[(cosx)^2

y=2sinx/根号下1-cos方x+cosx/根号下1-sin方x=2inx/|sinx|+cosx/|cosx|X在第一象限时,y=3X在第二象限时,y=1X在第三象限时,y=-3X在第四象限时,

2sin(x+π/4）+1≥0sin(x+π/4)≥-1/22kπ-π/6≤x+π/4≤2kπ+7π/62kπ-5π/12≤x≤2kπ+11π/12[2kπ-5π/12,2kπ+11π/12],k∈Z

y=1/根号下1-sin^2x=1/根号下cos²xcos²x≠0x≠kπ+π/2

dy/dx=sin(x^1/2)/y*x^(1/2)1/ydy=sin(x^1/2)/x^(1/2)dx=2d(sin(x^1/2))lny=2sin(x^1/2)+cy=c1*e^(2sin(x^1

复合求导,先把ln后面的式子看成整体f（x）,写成它的倒数,再乘以整体f(X)的导数

3sinx-2sin^2x-1>=0令sinx=tt范围为[-1,1]3t-2t^2-1>=0所以(t-1/2)(t-1)

X>=0根号里面的数大于等于0