ta có m=(1+1/3+1/5+1/7+.....+1/2019)-(1/2+1/4+......+1/2018)

=
(1+1/2+1/3+.......+1/2019)-(1/2+1/4+.....+1/2018)-(1/2-1/4-1/6-.....-1/2018)

=(1+1/2+1/3+.....+1/2019)-2.(1/2+1/4+.....+1/2018)

=(1+1/2+1/3+....+1/2019)-(1+1/2+.....+1/2019

=1/2010+1/2011+.....+1/2019=N

=>(m-n)^2=0^2=0

vậy( m-n)^2=0

Ta có: \(\left(1-\frac{1}{2^2}\right)\times\left(1-\frac{1}{3^2}\right)\times\left(1-\frac{1}{4^2}\right)\times...\times\left(1-\frac{1}{50^2}\right)\)

    \(=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{2499}{2500}\)

    \(=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times...\times\frac{49.51}{50.50}\)

    \(=\frac{1.2.3.....49}{2.3.4.....50}\times\frac{3.4.5.....51}{2.3.4.....50}\)

    \(=\frac{1}{50}\times\frac{51}{2}\)

    \(=\frac{51}{100}\)

• mọi người làm nhanh tui tích cho 

theo đề ra ta có :x/5=y/1=z/-2=>x/5=y/1=2z/-4=x+y-2z/5+1-(-4)=160/10=16

=>x/5=16;y/1=16;z/-2=16=>x=80;y=16;z=-32

vậy x=80;y=16;z=-32

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

\(\frac{x}{5}=\frac{y}{1}=\frac{z}{-2}=\frac{x+y-2z}{5+1+4}=\frac{160}{10}=16\)

\(x=16.5=80\)

\(y=16.1=16\)

\(z=16\left(-2\right)=-32\)

Ta có: \(\left(3x-2\right)^4=\left(3x-2\right)^6\)

    \(\Leftrightarrow\left(3x-2\right)^4-\left(3x-2\right)^6=0\)

    \(\Leftrightarrow\left(3x-2\right)^4\left[1-\left(3x-2\right)^2\right]=0\)

+ \(\left(3x-2\right)^4=0\)\(\Leftrightarrow\)\(3x-2=0\)\(\Leftrightarrow\)\(x=\frac{2}{3}\)

+ \(1-\left(3x-2\right)^2=0\)\(\Leftrightarrow\)\(\left(3x-2\right)^2=1\)

                                                \(\Leftrightarrow\)\(3x-2=\pm1\)

                                                \(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}\)

Vậy \(x=\frac{2}{3}\)hoặc \(x=\frac{1}{3}\)hoặc \(x=1\)

Ta có \(\frac{2010c-2011b}{2009}=\frac{2011a-2009c}{2010}=\frac{2009b-2010a}{2011}\)

=> \(\frac{2009.2010c-2009.2011b}{2009^2}=\frac{2010.2011a-2009.2010c}{2010^2}=\frac{2009.2011b-2010.2011a}{2011^2}\)

= \(\frac{2009.2010c-2009.2011b+2010.2011a-2009.2010c+2009.2011b-2010.2011a}{2009^2+2010^2+2011^2}\)= 0

=> \(\hept{\begin{cases}2010c-2011b=0\\2011a-2009c=0\\2009b-2010a=0\end{cases}}\Rightarrow\hept{\begin{cases}2010c=2011b\\2011a=2009c\\2009b=2010a\end{cases}}\Rightarrow\hept{\begin{cases}\frac{c}{2011}=\frac{b}{2010}\\\frac{a}{2009}=\frac{c}{2011}\\\frac{b}{2010}=\frac{a}{2009}\end{cases}}\)

=> \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}\)(đpcm)

Đặt \(\frac{a}{2018}=\frac{b}{2019}=\frac{c}{2020}=k\Rightarrow\hept{\begin{cases}a=2018k\\b=2019k\\c=2020k\end{cases}}\)

Khi đó 4(a - b)(b - c) = 4(2018k - 2019k)(2019k - 2020k)

= 4(-k).(-k) 

= 4k² (1)

Lại có (c - a)² = (2020k - 2018k)² = (2k)² = 4k² (2)

Từ (1)(2) => 4(a - b)(b - c) = (c - a)²

ta có

\(\frac{1}{97.95}=\frac{1}{2}.\left(\frac{1}{95}-\frac{1}{97}\right)\)

\(\frac{1}{95.93}=\frac{1}{2}.\left(\frac{1}{93}-\frac{1}{95}\right)\)

.....

\(\frac{1}{3.1}=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}\right)\)

.vì vậy \(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}=\)

\(\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}-\frac{1}{95}+\frac{1}{97}-\frac{1}{93}+\frac{1}{95}-\frac{1}{3}+\frac{1}{5}-1+\frac{1}{3}\right)\)

\(=\frac{1}{2}\left(\frac{2}{97}-\frac{1}{99}-1\right)\)

(hy vọng bạn không nhầm ở phần tử đầu tiên)