y=1 lg(x 1)-3
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4-log2x>=0且x>0∴00且5-3x>0∴1=
y'=(lg(1+cosx))'*(1+cosx)'=1/(1+cosx)*(-sinx)=-sinx/(1+cosx)再问:公式里(lgx)'=(1/x)lge的lg在这里的结果没有了?
x=[10^(y-1)+3]/2
令g(x)=x2ln(1+x1−x),x∈[-12,12],则g(-x)=x2ln(1−x1+x)=-g(x),即g(x)为奇函数,∴g(x)max+g(x)min=0,∵3+x2ln(1+x1−x)
(x-y)(x+3y)=2^2*xyx^2+2xy-3y^2=4xyx^2-2xy-3y^2=0(x+y)(x-3y)=0x=-y,x=3y由定义域x>0,y>0x=-y不成立x=3yx/y=3
通分,对数运算,结果x^2+y^2=1
当X1
lg(lgy)=lg(3x)+lg(3-x)有意义∴0<x<3∴lgy=lg3x*(3-x)∴y=10^(9x-3x^2),定义域为(0,3)(2)设U=-3X^2-9X=-3(x-3/2)^2+27
答:lg(x-3y)^2=lg4xy(x-3y)^2=4xyx^2-6xy+9y^2=4xyx^2-10xy+9y^2=0(x-y)(x-9y)=0x=yorx=9yy/x=1ory/x=1/9
∵y=1/lgx∴x>0lgx≠0∴x∈(0,1)∪(1,+∞)
2lg(x-3y)=lgx+lg(4y)lg(x-3y)²=lg(4xy)(x-3y)²=4xyx²-6xy+9y²-4xy=0x²-10xy+9y&
(1)∵lg(3x)+lgy=lg(x+y+1),∴lg(3xy)=lg(x+y+1),且x>0,y>0则3xy=x+y+1,∵3xy=x+y+1≥2xy+1,解得xy≥1,即xy≥1.即xy的最小值
lg(x-3)+lg(x-6)=1lg(x-3)(x-6)=1则,(x-3)(x-6)=10即x^2-9x+8=0(x-8)(x-1)=0所以x=8或者x=1
f(y)的定义域为[-1,2),则f(y-1)的定义域就表示-1≤x-1<2
∵x(n+1)=x²n+xn-1/4∴x(n+1)+1/2=x²n+xn+1/4=(xn+1/2)²两边取对数:lg[x(n+1)+1/2]=lg(xn+1/2)
lg(lgy)=lg(3x)+lg(3-x)=lg[3x(3-x)]∴lgy=3x(3-x)∴y=10^[3x(3-x)]=10^(9x-3x^2)=1000^(3x-x^2)∴f(x)=1000^(
答:1)y=f(x),lg(lgy)=lg(3x)+lg(3-x)=lg[(3x(3-x)]所以:lgy=3x(3-x)>0所以:y=e^(9x-3x^2),0再问:�������Ǹ�һ�ģ�����