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设F(x)=?log3(x+1),x>63x?6?1,x≤6,满足F(n...

f-1(-8/9)=n,f(x)过(n,-8/9) n>6时,log3(n+1)=8/9,n不存在 n

(1)原不等式可转化为:x>03?2k?1?x>0x(3?2k?1?x)≥22k?1即x>0x<3?2k?1x2?3?2k?1x+2k?1?2k≤0∴2k-1≤x≤2k(4分)∴f(k)=2k-(2k-1-1)=2k-1+1.(6′)(2)∵Sn=f(1)+f(2)+…+f(n)=20+21+…+2n-1+n=2n+n-1.(10′)(3)∵Tn=log22nlog22n...

(1)由4?x≥0x+1>02x?3≠0得x≤4x>?1x≠log23所以?1<x≤4且x≠log23所以原函数的定义域为(-1,log23)∪(log23,4](2)解:由1-log2(4x-5)≥0得log2(4x-5)≤1所以4x?5>04x?5≤2解得54<x≤74所以原函数的定义域为(54,74]

由对数的运算法则可知:log2(9x) + log2(x-1/3) = log2[9x(x-1/3)] 。 又根据对数的性质可知,1可以写成:1 = log2(2) 从而log2[9x(x-1/3)] = log2(2) 即:9x(x-1/3) = 2 9x^2 - 3x -2 = 0 十字交叉:(3x-2)(3x+1) = 0 x1=2/3 x2= -1 (有两个解...

我会,等下,给过程

∵f(x)=|log3x|,正实数m,n满足m<n,且f(m)=f(n),∴-log3m=log3n,∴mn=1.∵f(x)在区间[m,n2]上的最大值为2,函数f(x)在[m,1)上是减函数,在(1,n2]上是增函数,∴-log3m=2,或log3n2=2.若-log3m=2,则m=3-2=19,故n=9,n2=81,故...

(1)证明见解析(2)S n =n·2 n+3 (1)证明 f(a n )=4+(n-1)×2=2n+2,即log a a n ="2n+2, " 2分可得a n =a 2n+2 .∴ = = =a 2 (n≥2)为定值. 4分∴{a n }为等比数列. 6分(2)解 b n =a n f(a n )=a 2n+2 log a a 2n+2 =(2n+2)a 2n+2 .当a= 时,...

用符号语言比较简单: syms x f=log2(x); taylor(f, 4,2) %这个是n=4,x0=2展开,n、x0可以换的; >> ans = (x - 3)/(3*log(2)) - (x - 3)^2/(18*log(2)) + (x - 3)^3/(81*log(2)) - (x - 3)^4/(324*log(2)) + log(3)/log(2) >> pretty(ans) 2 ...

∵f(x)=log 3 (x-3),f(m)+f(3n)=2,∴ m-3>0 3n-3>0 lo g 3 (m-3)+lo g 3 (3n-3)=2 ,解得 m>3 n>1 3 m + 1 n =1 .∴m+n= (m+n)( 3 m + 1 n ) =4+ 3n m + m n ≥2 3n m × m n +4= 2 3 +4 ,当且仅当 3n m = m n ,m>3,n>1, 3 m +...

∵f(x)=log 2 (x-2),若实数m,n满足f(m)+f(2n)=3,m>2,n>1,∴log 2 (m-2)+log 2 (2n-2)=3,log 2 (m-2)2(n-1)=3,(m-2)2(n-1)=8,(m-2)(n-1)=4,∴ (m-2)(n-1) =2≤ m-2+n-1 2 = m+n-3 2 (当且仅当m-2=n-1=2时,取等...

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