Cho C = 1 + 3 1 + 3 2 + 3 3 + . . . + 3 11 . Chứng minh rằng:
a, C ⋮ 13
b, C ⋮ 40
Bài 1: Cho a,b,c >0 t/m: abc=1
CMR: \(\dfrac{1}{a^3+b^3+1}+\dfrac{1}{b^3+c^3+1}+\dfrac{1}{c^3+a^3+1}\le1\)
Bài 2: Cho a,b,c >0 t/m a+b+c=1
CMR: \(\dfrac{1+a}{1-a}+\dfrac{1+b}{1-b}+\dfrac{1+c}{1-c}\ge6\)
Bài 3: Cho a,b,c >0 t/m abc=1
CMR: \(\dfrac{ab}{a^4+b^4+ab}+\dfrac{bc}{b^4+c^4+bc}+\dfrac{ac}{c^4+a^4+ac}\le1\)
cho 1/a+1/b+1/c=1/a+b+c cmr 1/a^3+1/b^3+1/c^3=1/a^3+b^3+c^3
Lời giải:
Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)
\(\Leftrightarrow (a+b)\left[\frac{1}{ab}+\frac{1}{c(a+b+c)}\right]=0\)
\(\Leftrightarrow (a+b).\frac{c(a+b+c)+ab}{abc(a+b+c)}=0\Leftrightarrow (a+b).\frac{(c+a)(c+b)}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b)(b+c)(c+a)=0\)
\(\Rightarrow \left[\begin{matrix} a+b=0\\ b+c=0\\ c+a=0\end{matrix}\right.\)
Không mất tổng quát giả sử $a+b=0$
$\Rightarrow$
$\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{(-b)^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}$
\(\frac{1}{a^3+b^3+c^3}=\frac{1}{(-b)^3+b^3+c^3}=\frac{1}{c^3}\)
\(\Rightarrow \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3}\) (đpcm)
cho 3 số thực dương sao cho abc=1 c/m 1/a^3(b+c)+1/b^3(a+c)+1/c^3(a+b)>=3/2
a,Cho B = 1/2+1/2^2+1/2^3+...+1/2^99. So sánh B với 1
b, Cho C = 1/3+(1/3)^2+(1/3)^2+(1/3)^3+...+(1/3)^99. CMR C < 1/2
cho a;b;c> 0 a.b.c=1 .C/M (1/a^3+b^3+1 + 1/b^3+c^3+1 +1/a^3 +c^3+1)nhỏ hơn hoặc bằng 1
dùng bdt phụ nhé a^3 + b^3 >= ab(a+b)
a) cho B = 1/2 + 1/2^2 + 1/2^3 +....+1/2^99. só sánh B với 1
b) cho C = 1/3 +(1/3)^2 + (1/3)^2 + (1/3)^3 + ..... + (1/3)^99. CMR C<1/2
a,Cho B = 1/2+1/2^2+1/2^3+...+1/2^99. So sánh B với 1
b, Cho C = 1/3+(1/3)^2+(1/3)^2+(1/3)^3+...+(1/3)^99. CMR C < 1/2
ta có: 2B=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)
B=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+..+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
=>2B-B=\(1-\frac{1}{2^{99}}\)
mà 1/2^99>0 nên B<1 (đpcm)
Cho C = 1/3 +(1/3)^2b + (1/3)^2 + (1/3)^3 + ... (1/3)^99
CMR C < 1/2
Cho C = 1/3 + 1/3^2 + 1/3^2 + 1/3^3 + ... + 1/3^99
CMR C < 1/2
Cho C = 1 + 1/3 + 1/3^2+ 1/3^3+ ... + 1/3^2012 cm rằng C < 3/2
Các bạn giúp mình nhanh với nhá mình cần gấp lắm T^T
\(3C=3+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2013}}.\)
\(\Rightarrow3C-C=\frac{1}{3^{2013}}-3\)
\(\Rightarrow C=\frac{\frac{1}{3^{2013}}-3}{2}\le\frac{3}{2}\)
Study well