Ta có \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(\Rightarrow\frac{1}{3}.B=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)

\(\Rightarrow B-\frac{1}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)

\(\frac{2}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)

\(B=\left(\frac{1}{3}-\frac{1}{3^{2006}}\right):\frac{2}{3}\)

\(B=\frac{1}{3}:\frac{2}{3}-\frac{1}{3^{2006}}:\frac{2}{3}=\frac{1}{2}-\frac{1}{2.3^{2005}}< \frac{1}{2}\)

ta có      \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}.\)

      \(\Rightarrow3B=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

    \(\Leftrightarrow3B-B=1+\frac{1}{3}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^2}+...+\frac{1}{3^{2004}}-\frac{1}{3^{2004}}-\frac{1}{3^{2005}}\)

 \(\Leftrightarrow2B=1-\frac{1}{3^{2005}}\)   \(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\left(đpcm\right)\)

Có : 

3B = 1  +1/3 + 1/3^2 + ...... + 1/3^2004

2B = 3B - B = ( 1 + 1/3 + 1/3^2 + ....... + 1/3^2004 ) - ( 1/3 + 1/3^2 + ...... + 1/3^2004 )

     = 1 - 1/3^2004 < 1

=> B < 1/2

Tk mk nha

Bạn tham khảo nhé :) 

Ta có : 

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\)

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\) ( loại bỏ các phân số đối nhau ) 

\(2B=1-\frac{1}{3^{2005}}< 1\)

\(B< \frac{1}{2}\)

Vậy \(B< \frac{1}{2}\)

Chúc bạn học tốt ~

Với \(a,b,c\ne0\); \(a+b+c\ne0\) , ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)=abc\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+c\left(ab+bc+ca\right)=abc\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+abc+bc^2+c^2a=abc\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+bc^2+c^2a=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca\right)+c^2\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

Không mất tính tổng quát, ta lấy \(a=-b\), ta có:

\(\frac{1}{a^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{\left(-b\right)^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}\)

\(=\frac{-1}{b^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{c^{2005}}\) (1)

Ta có:\(\frac{1}{a^{2005}+b^{2005}+c^{2005}}=\frac{1}{\left(-b\right)^{2005}+b^{2005}+c^{2005}}\)

\(=\frac{1}{-b^{2005}+b^{2005}+c^{2005}}=\frac{1}{c^{2005}}\) (2)

Từ (1), (2), suy ra \(\frac{1}{a^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{a^{2005}+b^{2005}+c^{2005}}\)

Cái chỗ không mất tính tổng quát đấy, là do a, b, c bình đẳng nhau.

1. Ta có:

\(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( Nếu a, b ≥ 0)

=> \(a-2\sqrt{ab}+b\ge0\)

=> \(\left(a-2\sqrt{ab}+b\right)+2\sqrt{ab}\ge0+2\sqrt{ab}\)

=> \(a+b\ge2\sqrt{ab}\) => \(\frac{\left(a+b\right)}{2}\ge\frac{2\sqrt{ab}}{2}\)

=> \(\frac{\left(a+b\right)}{2}\ge\sqrt{ab}\);

(Dấu "=" xảy ra khi \(\sqrt{a}-\sqrt{b}=0\) => a = b)

1. BĐT \(\Leftrightarrow a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

2. BĐT \(\Leftrightarrow\frac{a+b}{2}\ge\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{4}\)

\(\Leftrightarrow2\left(a+b\right)\ge a+2\sqrt{ab}+b\)

\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( luôn đúng )

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

3. Ta có: \(M=\frac{2}{\sqrt{1\cdot2005}}+\frac{2}{\sqrt{2\cdot2004}}+...+\frac{2}{\sqrt{1003\cdot1003}}\)

Áp dụng BĐT Cô-si:

\(\sqrt{1\cdot2005}\le\frac{1+2005}{2}=1003\)

Do dấu "=" không xảy ra nên \(\sqrt{1\cdot2005}< 1003\)

Khi đó: \(\frac{2}{\sqrt{1\cdot2005}}>\frac{2}{1003}\)

Chứng minh tương tự với các phân thức còn lại rồi cộng vế ta được :

\(M>\frac{2006}{1003}>\frac{2005}{1003}\) ( đpcm )

Em có cách khác ở bài 2 nè:) Nhưng thôi làm 2 bài luôn bài 3 ý tưởng y hệt hà..

Bài 1: BĐT \(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\left(true\right)\)

Đẳng thức xảy ra khi a = b

Bài 2: BĐT trên là thuần nhất (hay đồng bậc gì ấy) nên ta chuẩn hóa a + b =2.

Cần chứng minh: \(1\ge\frac{\sqrt{a}+\sqrt{b}}{2}\)

Thật vậy theo Cô si: \(RHS\left(VP\right)=\frac{\sqrt{1.a}+\sqrt{1.b}}{2}\le\frac{\frac{a+1}{2}+\frac{b+1}{2}}{2}=\frac{a+b+2}{4}=1=LHS\left(VT\right)\)

Ta có đpcm. True?

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}} \)
\(2B=1-\frac{1}{3^{2004}}\)
\(B=\frac{1}{2}-\frac{1}{2\cdot3^{2004}}\)
Do đó B<\(\frac{1}{2}\)
chúc thành công

Ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}a+b=0\\b+c=0\\c+a=0\end{cases}}\)

Với \(a+b=0\)

Thì \(\hept{\begin{cases}\frac{1}{a^{2005}}+\frac{1}{b^{2005}}+\frac{1}{c^{2005}}=\frac{1}{c^{2005}}\\\frac{1}{a^{2005}+b^{2005}+c^{2005}}=\frac{1}{c^{2005}}\end{cases}}\)

Tương tự cho 2 trường hợp còn lại ta có ĐPCM

Có B=\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+...+\(\frac{1}{3^{2004}}\)+\(\frac{1}{3^{2005}}\)

=>3B=3.(\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\))

=>3B=1+\(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)

=>3B-B=(1+\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\))-(\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\))

=>2B=\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-....-\frac{1}{3^{2004}}-\frac{1}{3^{2005}}\)

=>2B=1-\(\frac{1}{3^{2005}}\)

=>B=(\(1-\frac{1}{3^{2005}}\)):2

Mà \(\left(1-\frac{1}{3^{2005}}\right)< \frac{1}{2}\)=>\(\left(1-\frac{1}{3^{2005}}\right):2< \frac{1}{2}\)

=>B<\(\frac{1}{2}\)(đpcm)

1.

A=19^5^1^8^9^0+2^9^1^9^6^9

Ta luôn có 1^a=1 với a là số nguyên dương

=>19^5^1^8^9^0=19⁵ và 2^9^1^9^6^9=2⁹

=>A=19⁵+2⁹=(19²)².19+(2⁴)².2=(...1)².19+(...6)².2=...1.19+...6.2=...1

Vậy A có tận cung là 1.

2.

B=1/3+1/3²+...+1/3²⁰⁰⁵

3B=1+1/3+1/3²+...+1/3²⁰⁰⁴

3B-B=1-1/3²⁰⁰⁵

2B=1-1/3²⁰⁰⁵<1

=>2B<1=>B<1/2

Vậy B<1/2.

.

.

1) Ta có:

\(19^{5^{1^{8^{9^0}}}}+2^{9^{1^{9^{6^9}}}}=19^{5^1}+2^{9^1}\)

Mà 19⁵=19⁴⁺¹=...1.19=...19

      2⁹=2^2.4+1=...6 .2=...2

=>A=...19 + ...2= ...1

Vậy A có chữ số tận cùng là 1

1. 

\(A=19^{5^{1^{8^{9^0}}}}+2^{9^{1^{9^{6^9}}}}\)

Ta có

\(19^{5^{1^{8^{9^0}}}}=0\)

\(2^{9^{1^{9^{6^9}}}}=....6\)

=> \(A=19^{5^{1^{8^{9^0}}}}+2^{9^{1^{9^{6^9}}}}=0+...6=...6\)

=> A có chữ số tận cùng là 6

Có : 

3B = 1 + 1/3 + 1/3^2 + .... + 1/3^2004

2B = 3B - B = ( 1 + 1/3 + 1/3^2 + ..... + 1/3^2004 ) - ( 1/3 + 1/3^2 + 1/3^3 + ..... + 1/3^2005 )

                  = 1 - 1/3^2005 < 1

=> B < 1 : 2 = 1/2

=> ĐPCM

Tk mk nha

\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)

\(\Rightarrow3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2B=1-\frac{1}{3^{2005}}< 1\)

\(\Rightarrow B< \frac{1}{2}\)

Kết bạn LQM ko

Tên 

$ YASUO $