y+sin+3x+1+求y

巫尝眨3148已知cos(x - y/2)= - 1/9,sin(x/2 - y)=2/3,0<x<π,0<y<π/2求cos(x+y)的值 -
毕贤映13425458218 ______ cos(x-y/2)=-1/9,sin(x/2-y)=2/3,0<x<π,0<y<π/2求cos(x+y) 因为cos(x-y/2)=-1/9 0<x<π,0<y<π 所以 π/2<(x-y/2)<π sin(x-y/2)=4*√5/9 因为sin(x/2-y)=2/3,0<x<π,0<y<π/2 所以 0<(x-y/2)<π/2 cos(x/2-y)=√5/3 cos[(x-y/2)-(x/2-y)]=cos(x-y/2)*cos(x/2-y)+sin(x-y/2)...

巫尝眨3148已知实数(x,y)满足x方+y方 - 2x+4y - 20=0则x方+y方的最大值和最小值 -
毕贤映13425458218 ______ x²+y²-2x+4y-20=0 (x²-2x+1)+(y²+4y+4)=20+1+4 (x-1)²+(y+2)²=5² 令x=1+5cosa,y=-2+5sina,则有 x²+y² =(1+5cosa)²+(-2+5sina)² =1+10cosa+25cos²a+4-20sina+25sin²a =1+4+25(cos²a+sin²a)-(20sina-10cosa) =30-10(...

巫尝眨3148x=2+3cosθ,y= - 1+3sinθ; 求:①X+Y的最值;②X^2+Y^2最值; ③求(X,Y)到3X+4Y - 1=0的距离的最值; -
毕贤映13425458218 ______[答案] 1)X+Y=1+3cosθ+3sinθ 用辅助角 =1+3根号2 sin(θ+π/4) 所以 最大值为1+3根号2,此时sin(θ+π/4)=1,θ=2kπ+π/4 最小值为1-3根号2,此时sin(θ+π/4)=-1,θ=2kπ-3π/4 2)X^2+Y^2=14+12cosθ-6sinθ 继续辅助角,tan t=6/12=1/2 =14+6根号5 cos(θ+t) 最大值为...

巫尝眨3148设函数y=f(x)由方程e∧y+sin(x+y)=1决定,求二阶导数 -
毕贤映13425458218 ______ 两边对x求导:y'e^y+(1+y')cos(x+y)=0, 1)这里可得到y'=-cos(x+y)/[e^y+cos(x+y)]再对1)求导:y＂e^y+(y')^2e^y+y＂co...

巫尝眨3148若实数x,y满足x的平方+y的平方 - 2x+4y=0,求x - 2y的最大值 -
毕贤映13425458218 ______ 方法一:x²+y²-2x+4y=0 →(x-1)²+(y+2)²=(√5)².设x-1=√5cosθ,y+2=√5sinθ,则 x-2y=(1+√5cosθ)-2(-2+√5sinθ) =5+√5(cosθ-2sinθ) =5+5cos(θ+φ) (tanφ=2) ∴cos(θ+φ)=1时,所求最大值为:(x-2y)|max=10; cos(θ+φ)=-1时,所求最小值为...

巫尝眨3148已知x^2+y^2 - 2x - 4y - 4=0,求x+y取值范围 -
毕贤映13425458218 ______ x^2+y^2-2x-4y-4=0(x-1)^2+(y-2)^2=9 令x=1+3cosa,y=2+3sina x+y=3(1+cosa+sina) =3+3(sina+cosa) =3+3√2sin(a+π/4) 因为-1≤sin(a+π/4)≤1 所以3-3√2≤x+y≤3+3√2

巫尝眨3148设2y+cos(x+y)=0,求dy/dx -
毕贤映13425458218 ______ 两边对x求导得2y'-sin(x+y)+1+y'=03y'=sin(x+y)-1 y'=[sin(x+y)-1]/3 dy=[sin(x+y)-1]/3*dx

巫尝眨3148已知,Y=(x+1)/(2x+1)求1.Y的取值范围2.当x≥1时,Y的取值范围,急急急! -
毕贤映13425458218 ______ 解:(1)y=[1/2·(2x+1)+1/2]/(2x+1)=1/2+1/2(2x+1)≠1/2(2)∵y=(x+1)/(2x+1) ∴(2y-1)x=1-y x=(1-y)/(2y-1) ∵x≥1 ∴(1-y)/(2y-1)≥1 ∴(-3y+2)/(2y-1)≥0 ∴1/2<y≤2/3

巫尝眨3148不等式习题已知正数x、y、z满足:xy+yz+zx=1.求证:x
毕贤映13425458218 ______ 设x=tan(A/2),y=tan(B/2),z=tan(C/2),A+B+C=π. 则原不等式等价于sin(A/)+sin(B/2)+sin(C/2)≤3/2. 而依Jensen不等式得 sin(A/2)+sin(B/2)+sin(C/2)≤3sin[(A+B+C)/6]=3sin(π/6)=3/2, 故原不等式得证.

巫尝眨3148cos(x^2+y)(2x+y')=y+xy'求y'=什么?要详细过程!谢谢! -
毕贤映13425458218 ______ 郭敦顒回答:∵cos(x^2+y)(2x+y')=y+xy'=y+x ∴2xcos(x^2+y)+y'(x^2+y)=y+x y'(x^2+y)=y+x-2xcos(x^2+y) ∴y′=[y+x-2xcos(x^2+y)]/(x^2+y) 或∵cos(x^2+y)(2x+y')=y+xy' ∴cos(x^2+y)(2x+1)-y=xy' ∴y′=[cos(x^2+y)(2x+1)-y]/x