pạn -1 vào mỗi phân số là xong. Rùi ra x\(\frac{x-2015}{1986}\)+\(\frac{x-2015}{1988}\)+ \(\frac{x-2015}{1990}\)+...+\(\frac{x-2015}{x1996}\)-\(\frac{x-2015}{29}\)-\(\frac{x-2015}{27}\)-...\(\frac{x-2015}{19}\)=0

<=>(x-2015)(\(\frac{1}{1986}\)+\(\frac{1}{1988}\)+... -\(\frac{1}{19}\))=0...(mà \(\frac{1}{1986}\)+...- \(\frac{1}{19}\) khác 0)

=>x-2015=0

<=> x=2015

\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)

\(\Rightarrow\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}+1\)

\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)

\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)

\(\Rightarrow\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)

Mà \(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\ne0\)

\(\Rightarrow x+59=0\)

\(\Rightarrow x=-59\)

ko vt lại đề 

=> (x-1)/15 -133 + ( x-5)/10 -197 + x/5 -392 -2019 =0

=> (x-1)/15 + (x-5)/10 + x/5 - 2741=0

=> (2x-2)/30 + (3x-15)/30 + 6x/30 =2741

=> ( 2x-2+3x-15+6x)/30 =2741

=> 11x-17=82230

=> 11x= 82247

=> x= 7477

vì 7477 là số nguyên => nghiệm của phương trình là số nguyên 

Vậy....

hướng dẫn thôi nhé 

Có: \(\left(\frac{16}{\sqrt{x-1996}}+\sqrt{x-1996}\right)+\left(\frac{1}{\sqrt{y-2008}}+\sqrt{y-2008}\right)\)

\(\ge2\sqrt{\frac{16}{\sqrt{x-1996}}\sqrt{x-1996}}+2\sqrt{\frac{1}{\sqrt{y-2008}}\sqrt{y-2008}}=8+2=10\)

\(\Leftrightarrow\)\(\frac{16}{\sqrt{x-1996}}+\frac{1}{\sqrt{y-2008}}\ge10-\left(\sqrt{x-1996}+\sqrt{y-2008}\right)\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\frac{16}{\sqrt{x-1996}}=\sqrt{x-1996}\\\frac{1}{\sqrt{y-2008}}=\sqrt{y-2008}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2012\\y=2009\end{cases}}\)

a, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1998}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)

=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1998}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1998}+\frac{x-2004}{2000}\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)

=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}=0\right)\)

=> \(x-2004=0\)

=> \(x=2004\)

Vậy phương trình có nghiệm là x = 2004 .

b, Ta có : \(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)

=> \(\frac{x-85}{15}-1+\frac{x-74}{13}-2+\frac{x-67}{11}-3+\frac{x-64}{9}-4=10-1-2-3-4=0\)

=> \(\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)

=> \(\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)

=> \(x-100=0\)

=> \(x=100\)

Vậy phương trình có nghiệm là x = 100 .