Đặt \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=k\)

\(\Rightarrow a=2003k;b=2004k;c=2005k\)

Thay a = 2003k, b = 2004k, c = 2005k vào 4(a - b)(b - c), ta có:

4(2003k - 2004k)(2004k - 2005k)

= 4(-k)(-k)

= 4k²

Thay a = 2003k, b = 2004k, c = 2005k vào (c - a)², ta có:

(2005k - 2003k)² = (2k)² = 4k²

Vì 4k² = 4k² nên 4(a - b)(b - c) = (c - a)²

Vậy với \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}\)thì \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\) 

\(\dfrac{a+2003}{a-2003}=\dfrac{b-2004}{b+2004}\)

\(\Leftrightarrow\left(a+2003\right)\left(b+2004\right)=\left(a-2003\right)\left(b-2004\right)\)

\(\Leftrightarrow ab+2004a+2003a+2003\cdot2004=ab-2004a-2003a+2003\cdot2004\)

\(\Leftrightarrow4008a=4006b\)

=>a/b=2003/2004

hay a/2003=b/2004

cm cái j

Giải:

Đặt \(\dfrac{a}{2003}=\dfrac{b}{2004}=\dfrac{c}{2005}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2003k\\b=2004k\\c=2005k\end{matrix}\right.\)

Ta có:

\(4\left(a-b\right)\left(b-c\right)\)

\(=4\left(2003k-2004k\right)\left(2004k-2005k\right)\)

\(=4.\left(-k\right)\left(-k\right)\)

\(=4.k^2\) (1)

Lại có:

\(\left(c-a\right)^2\)

\(=\left(2005k-2003k\right)^2\)

\(=\left(2k\right)^2\)

\(=4k^2\) (2)

Từ (1) và (2) \(\Rightarrow4\left(a-b\right)\left(a+b\right)=\left(c-a\right)^2\)

\(\Rightarrowđpcm\).

Chúc bạn học tốt!!!

Đặt:

\(\dfrac{a}{2003}=\dfrac{b}{2004}=\dfrac{c}{2005}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=2003k\\b=2004k\\c=2005k\end{matrix}\right.\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(2003k-2004k\right)\left(2004k-2005k\right)\)

\(=4.-k.-k=4k^2\)

\(\left(c-a\right)^2=\left(2005k-2003k\right)^2=2k^2=4k^2\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)

\(\rightarrowđpcm\)

Bài này mk làm rồi bạn, vào câu hỏi tương tự nhé!

Đặt \(\frac{a}{2002}=\frac{b}{2003}=\frac{c}{2004}=k\)

\(\Rightarrow\hept{\begin{cases}a=2002k\\b=2003k\\c=2004k\end{cases}}\)

\(VT=4\left(a-b\right)\left(b-c\right)=4\left(2002k-2003k\right)\left(2003k-2004k\right)=4\left(-1k\right)\left(-1k\right)=4k^2\)

\(VP=\left(c-a\right)^2=\left(2004k-2002k\right)^2=\left(2k\right)^2=4k^2\)

\(\Rightarrow VT=VP\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\left(đpcm\right)\)
 

4) Ta có :\(\frac{a+1}{2}=\frac{b-1}{3}=\frac{c+2}{4}=\frac{a+b+c+2}{2a+5}=\frac{a+b+c+1-1+2}{2+3+4}=\frac{a+b+c+2}{9}\)(1)

=> 2a + 5 = 9

=> 2a = 4

=> a = 2

Thay a vào (1) ta có : 

\(\frac{b-1}{3}=\frac{c+2}{4}=\frac{3}{2}\)

=> \(\hept{\begin{cases}\frac{b-1}{3}=\frac{3}{2}\\\frac{c+2}{4}=\frac{3}{2}\end{cases}}\Rightarrow\hept{\begin{cases}2\left(b-1\right)=9\\2\left(c+2\right)=12\end{cases}}\Rightarrow\hept{\begin{cases}2b-2=9\\2c+4=12\end{cases}}\Rightarrow\hept{\begin{cases}2b=11\\2c=8\end{cases}\Rightarrow\hept{\begin{cases}b=5,5\\c=4\end{cases}}}\)

Vậy a = 2 ; b = 5,5 ; c = 4

5) Đặt \(\frac{a}{2002}=\frac{b}{2003}=\frac{c}{2004}=k\)

=> \(\hept{\begin{cases}a=2002k\\b=2003k\\c=2004k\end{cases}}\)

4(a - b)(b - c) = (c - a)²

=> 4(2002k - 2003k)(2003k - 2004k) = (2002k - 2004k)²

=> 4(-k)(-k) = (-2k)²

=> (-2)²(-k)² = (-2k)²

=> 2²k² = (2k)²

=> (2k)² = (2k)²

=> 4(a - b)(b - c) = (c - a)² (đpcm)

Bài 4:

\(\frac{a+1}{2}=\frac{b-1}{3}=\frac{c+2}{4}=\frac{a+b+c+2}{2a+5}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\frac{a+1}{2}=\frac{b-1}{3}=\frac{c+2}{4}=\frac{a+1+b-1+c+2}{2+3+4}=\frac{a+b+c+2}{9}\)

\(\Rightarrow2a+5=9\Rightarrow a=2\)

Lại có: \(\frac{a+1}{2}=\frac{3}{2}\)\(\Rightarrow\frac{b-1}{3}=\frac{3}{2}\Leftrightarrow2\left(b-1\right)=9\Leftrightarrow b=\frac{11}{2}\)

\(\frac{c+2}{4}=\frac{3}{2}\Leftrightarrow2\left(c+2\right)=12\Leftrightarrow c=4\)

Vậy ...

dat \(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=k\)

suy ra \(\hept{\begin{cases}a=2003k\\b=2004k\\c=2005k\end{cases}}\)

4.(a-b).(b-c)=4.(2003k-2004k).(2004k-2005k)=4k^2

(c-a)^2=(2005k-2003k)^2=4k^2

xong roi do cho minh dung nhe!

 Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=\frac{a-b}{2003-2004}=\frac{b-c}{2004-2005}=\frac{c-a}{2005-2003}\)

\(\Rightarrow-\left(a-b\right)=-\left(b-c\right)=\frac{c-a}{2}\)

Thay vào \(4\left(a-b\right)\left(b-c\right)\), ta được :

\(4\left(a-b\right)\left(b-c\right)=4\left(-\frac{c-a}{2}\right)\left(-\frac{c-a}{2}\right)\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left[\frac{\left(c-a\right)^2}{4}\right]\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)( điều phải chứng minh ) 

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\frac{a}{2003}=\frac{b}{2004}=\frac{c}{2005}=\frac{a-b}{2003-2004}=\frac{b-c}{2004-2005}=\frac{c-a}{2005-2003}2003a​=2004b​=2005c​=2003−2004a−b​=2004−2005b−c​=2005−2003c−a​

\Rightarrow-\left(a-b\right)=-\left(b-c\right)=\frac{c-a}{2}⇒−(a−b)=−(b−c)=2c−a​

Thay vào 4\left(a-b\right)\left(b-c\right)4(a−b)(b−c), ta được :

4\left(a-b\right)\left(b-c\right)=4\left(-\frac{c-a}{2}\right)\left(-\frac{c-a}{2}\right)4(a−b)(b−c)=4(−2c−a​)(−2c−a​)

\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left[\frac{\left(c-a\right)^2}{4}\right]⇒4(a−b)(b−c)=4[4(c−a)2​]

\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2⇒4(a−b)(b−c)=(c−a)2( điều phải chứng minh ) 

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

=>2B=1-1/3^2005

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

=>\(B=\dfrac{3^{2005}-1}{3^{2005}\cdot2}< \dfrac{1}{2}\)

         B  =      \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\) +........+ \(\dfrac{1}{3^{2024}}\)+ \(\dfrac{1}{3^{2005}}\)

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

    3B -B  = 1 - \(\dfrac{1}{3^{2005}}\)

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

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

            B  =  \(\dfrac{1}{2}\) - \(\dfrac{1}{2.3^{2005}}\) < \(\dfrac{1}{2}\) (đpcm)