求Lg?=9.3
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1-x>0且1+x>0;定义域:x∈(-1,1)f(x)=lg(1-x)-lg(1+x)=lg[(1-x)/(1+x)](1)f(-x)=lg[(1+x)/(1-x)]=-lg[(1-x)/(1+x)
答:f(x)=lg(3+x)+lg(3-x),-30的单调递减区间[0,3)就是f(x)的单调减区间
lg(lgy)=lg3x+lg(3-x)=lg3x*(3-x)所以lgy=3x*(3-x)y=10的3x*(3-x)次y=(1/10)^(3x^2-9x)
∵lg(x-y)+lg(x+2y)=lg2+lgx+lgy,∴lg(x-y)(x+2y)=lg2xy.∴(x-y)(x+2y)=2xy,即(x-2y)(x+y)=0.再由x、y都是正数可得x+y≠0,
解lg(3x-1)+lg(12-x)=2lg(3x-1)×(12-x)=lg100∴(3x-1)(12-x)=100即36x-3x²-12+x=100即3x²-37x+112=0∴
(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
lg(27)/lg(12)=alg(2×3)/lg(3)=1/a(2lg2+lg3)/3lg3=1/a(2lg2)/(3lg3)=(1/a)-1/3(lg2)/(lg3)=(3/2a)-1/2
lg(x+2y)+lg(x-y)=lg2+lgx+lgylg(x+2y)(x-y)=lg2xy(x+2y)(x-y)=2xyx^2+xy-2y^2=2xyx^2-xy-2y^2=0(x-2y)(x+y
sin2x>0;∴0+2kπ<2x<π/2+2kπ(k∈Z)0+kπ
解由题知x>0,故函数的定义域为(0,正无穷大)令t=lgx,则t属于R则原函数变为y=-t^2+6t=-(t-3)^2+9故当t=3时,y有最大值9故原函数的值域为(负无穷大,9].
lg(x-y)+lg(x+y)=lg2+lgx+lgy左边=lg[(x-y)(x+y)]=lg[x^2-y^2]右边=lg(2xy)左边=右边所以lg[x^2-y^2]=lg(2xy)x^2-y^2=
真数>01+2x>0x>-1/21-3x>0x
不是lg2*lg5=lg(10/5)*lg5=(lg10-lg5)*lg5=(1-lg5)*lg5=lg5-(lg5)^2
真数大于0y=-2(lg²x-3lgx)=-2(lgx-3/2)²+9/2所以定义域是(0,+∞)值域[[-∞,9/2]
2lg(x-3y)=lgx+lg(4y)lg(x-3y)²=lg(4xy)(x-3y)²=4xyx²-6xy+9y²-4xy=0x²-10xy+9y&
Ac=20log(Ad)=20log(2.5)=20[log(10/4)]=20[log10-log4]=20[1-2log2]=20-40*log2...ans
LGx=LGa+LGb=LG(ab),由于LG函数是单调递增的,所以x=ab.
1,当p1时,定义域为(1,p)原式可化简为lg(x+1)(p-x)①1