y=mcos(-2x π 6) n在0,π 3上值域是-1,1求mn
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f(x)=sin2x+23cos2x−3=2sin(2x+π3)∵x1∈[0,π4]∴2x1+π3∈[π3,5π6]∴f(x1)∈[1,2]∵x2∈[0,π4]∴2x2−π6∈[−π6,π3]∴cos
y=(m+1)x-(n^2-2n-5)在y轴上的截距为-3,即-(n^2-2n-5)=-3.n=4,n=-2.且y的值随x增大而减小,m-1
2x(M+3x)=6x²y²+N2xM+6x²=6x²y²+N所以N=6x²,且2xM=6x²y²,即M=3xy&sup
将M与N中的两方程联立得:y=x+3y=−2x+6,解得:x=1y=4,∴M∩N={(1,4)}.故答案为:{(1,4)}
这题很难吗?只要知道2(n/3)^n*(n+1)=[n/(n+1)]^n*[(n+1)/3]^n*(n+1)>1/3*[(n+1)/3]^n*(n+1)=[(n+1)/3]^{n+1}(n+1)!=n
cos(x+π/4)=√2/2(cosx-sinx)令cosx-sinx=t-√2
解析2*(M+3x)=2*m+6x=6x²y²+N两边比较2*m=6x²y²m=3x²y²N=6x再问:我没看明白。请问这几个算式都是为什么
3x(M-5x)=6x^2y^3+N,3x*M-15x^2=6x^2y^3+NM=2xy^3N=-15x^2
由于2x(M+3x)=6x^2y^2+N,所以2xM+6x^2=6x^2y^2+N比较可知2xM=6x^2y^2,6x^2=N所以M=3xy^2,N=6x^2
x^6n+y^6n*x^4n=(x^3n)^2+(y^2n)^3*x^3n*x^n=4+54x^n
f(0)=-f(0)=>f(0)=0f(4m-2mcosθ)-f(2sin²θ+2)>f(0)=0=>f(4m-2mcosθ)>f(2sin²θ+2)0≤θ≤π/24m-2mcos
{x|x=(-1)^n,n∈N}={1,-1,1,-1,1,...}{y|y=-x^2+6,x∈N,y∈N}={6,5,2}{(x,y)|y=-x^2+6,x∈N,y∈N}={(0,6),(1,5),
M是指任意正实数.|x(n)|
(y^n)^3=y^3n=3所以原式=x^6n*y^6n=(x^2n)^3*(y^3n)^2=2^3*3^2=72
3xM=6x^2y^33x(-5x)=N所以M=2xy^3N=-15x^2
x^2n=2,x^6n=2^3=8,(y^n)^3=3,y^6n=3^2=9则(xy)^6n=x^6n*y^6n=8*9=72
x^2n=3x^6n=(x^2n)^3=3^3=27x^4n=(x^2n)^2=3^2=9y^9n=(y^3n)^3=2^3=82x^6n-x^4n×y^3m-y^9m=2*27-9*2-8=28
这两个均采用了配方法y=x²-4x+6=x²-4x+4+2=(x-2)²+2>=2y=-x²-2x+18=-(x²+2x+1)+19=-(x+1)
tan(α/2)=sin(α/2)/cos(α/2)=m/nnsin(α/2)=mcos(α/2)n2sin(α/2)cos(α/2)=m2cos²(α/2)=m(2cos²(α/
f(x)=mcos²x+√3msinxcosx+n=m/2(1+cos2x)+√3/2msin2x+n=m(√3/2sin2x+1/2cos2x)+m/2+n=msin(2x+π/6)+m/