196345−x​+196840−x​+197335−x​+197830−x​=−4
\left(\frac{45-x}{1963}+1\right)+\left(\frac{40-x}{1968}+1\right)+\left(\frac{35-x}{1973}+1\right)+\left(\frac{30-x}{1978}+1\right)=0(196345−x​+1)+(196840−x​+1)+(197335−x​+1)+(197830−x​+1)=0
\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=019632008−x​+19682008−x​+19732008−x​+19782008−x​=0
\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0(2008−x)(19631​+19681​+19731​+19781​)=0
=> 2008 - x = 0 ( vì 1/ 1963 + ... khác 0 )
=> x = 2008

Ta có : \(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)

\(\Leftrightarrow\frac{45-x}{1963}+1+\frac{40-x}{1968}+1+\frac{35-x}{1973}+1+\frac{30-x}{1978}=0\)

\(\Leftrightarrow\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=0\)

\(\Leftrightarrow\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

Vì \(\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)\ne0\)

Nên : 2008 - x = 0 

<=> x = 2008

Vậy x = 2008

d)\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}=-4\)

\(\Rightarrow\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+\frac{4\left(x+329\right)}{\left(x+329\right)}=0\)

\(\Rightarrow\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{\frac{1}{4}\cdot\left(x+329\right)}=0\)

\(\Rightarrow\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{\frac{1}{4}\left(x+329\right)}\right)=0\)

\(\Rightarrow x+329=0\).Do \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{\frac{1}{4}\left(x+329\right)}\ne0\)

=>x=-329

e)bn kiểm tra lại đề

a) \(x\left(x-2016\right)+2015\left(2016-x\right)=0\)

\(x\left(x-2016\right)-2015\left(x-2016\right)=0\)

\(\left(x-2015\right)\left(x-2016\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015 và x= 2016

b) \(-5x\left(x-15\right)+\left(15-x\right)=0\)

\(-5x\left(x-15\right)-\left(x-15\right)=0\)

\(\left(-5x-1\right)\left(x-15\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5 và x= 15

d) \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}=-4\)

\(\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0\)

\(\Leftrightarrow\left(\frac{55-x}{1963}+1\right)+\left(\frac{50-x}{1968}+1\right)+\left(\frac{45-x}{1973}+1\right)+\left(\frac{40-x}{1978}+1\right)=0\)

\(\Leftrightarrow\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0\)

\(\Leftrightarrow\left(2018-x\right).\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

\(\Leftrightarrow2018-x=0\)

\(\Leftrightarrow x=2018\)

Vậy \(x=2018\)

Dễ dàng :v

Có \(\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0\)

\(\Rightarrow\left(\frac{55-x}{1963}+1\right)+\left(\frac{50-x}{1968}+1\right)+\left(\frac{45-x}{1973}+1\right)+\left(\frac{40-x}{1978}+1\right)=0\)

\(\Rightarrow\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0\)

\(\Rightarrow\left(2018-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

Mà \(\Rightarrow\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)>0\Rightarrow2018-x=0\)

\(\Rightarrow x=2018-8=2018\)

Vậy x = 2018

a) Ta có: \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)

\(\Leftrightarrow\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)

\(\Leftrightarrow\frac{x-91-37}{37}+\frac{x-86-42}{42}+\frac{x-78-50}{50}+\frac{x-49-79}{79}=0\)

\(\Leftrightarrow\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)

\(\Leftrightarrow\left(x-128\right)\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)

Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)

nên x-128=0

hay x=128

Vậy: x=128

b) Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}-8=0\)

\(\Leftrightarrow\frac{x-29}{1970}-1+\frac{x-27}{1972}-1+\frac{x-25}{1974}-1+\frac{x-23}{1976}-1+\frac{x-1970}{29}-1+\frac{x-1972}{27}-1+\frac{x-1974}{25}-1+\frac{x-1976}{23}-1=0\)

\(\Leftrightarrow\frac{x-29-1970}{1970}+\frac{x-27-1972}{1972}+\frac{x-25-1974}{1974}+\frac{x-23-1976}{1976}+\frac{x-1970-29}{29}+\frac{x-1972-27}{27}+\frac{x-1974-25}{25}+\frac{x-1976-23}{23}=0\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}\right)=0\)

Vì \(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}>0\)

nên x-1999=0

hay x=1999

Vậy: x=1999

a) Ta có \(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}\)=4

<=>\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}-4=0\)

<=>\(\frac{x-91}{37}-1+\frac{x-86}{42}-1+\frac{x-78}{50}-1+\frac{x-49}{79}-1=0\)

<=>\(\frac{x-128}{37}+\frac{x-128}{42}+\frac{x-128}{50}+\frac{x-128}{79}=0\)

<=>(x-128)\(\left(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\right)=0\)

Vì \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}>0\)=>x-128=0<=>x=128

b)Tương tự

<=>x-128=0

<=>x=128

Chú ý \(\frac{1}{37}+\frac{1}{42}+\frac{1}{50}+\frac{1}{79}\)>0

b)tương tự

QĐMS :

X x 40/1200 + X x 30/1200 = 35 x 300/4 x 300 = 10500/1200

= X x 40 + X x 30/1200 = 10500/1200

= X x 70/1200 = 10500/1200

= X = 10500 : 70

= X = 150

\(x=150\)

\(\frac{x}{30}+\frac{x}{40}=\frac{35}{4}\)

\(\frac{4x}{120}+\frac{3x}{120}=\frac{1050}{120}\)

\(\Rightarrow4x+3x=1050\)

\(\Rightarrow7x=1050\)

\(\Rightarrow x=1050:7\)

\(\Rightarrow x=150\)

b)  \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)

\(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)=\left(\frac{x-95}{9}-1\right)\)\(+\left(\frac{x-93}{11}-1\right)\)

\(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)

\(\Leftrightarrow\left(x-104\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)

Mà  \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\ne0\)

\(\Rightarrow x-104=0\)

\(\Leftrightarrow x=104\)

Vậy ....

a)  \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)

\(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)\)\(+\left(\frac{x+1969}{69}-1\right)\)

\(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)

\(\Leftrightarrow\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)

Mà  \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\)

\(\Rightarrow x+1900=0\)

\(\Leftrightarrow x=-1900\)

Vậy ...

c)  \(\frac{x+2008}{10}+\frac{x+2010}{9}=\frac{x+2012}{8}+\frac{x+2014}{7}\)

\(\Leftrightarrow\left(\frac{x+2008}{10}+2\right)+\left(\frac{x+2010}{9}+2\right)\)\(=\left(\frac{x+2012}{8}+2\right)+\left(\frac{x+2014}{7}+2\right)\)

\(\Leftrightarrow\frac{x+2028}{10}+\frac{x+2028}{9}-\frac{x+2028}{8}-\frac{x+2028}{7}=0\)

\(\Leftrightarrow\left(x+2028\right)\left(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\right)=0\)

Mà  \(\frac{1}{10}+\frac{1}{9}-\frac{1}{8}-\frac{1}{7}\ne0\)

\(\Rightarrow x+2028=0\)

\(\Leftrightarrow x=-2028\)

Vậy ...

tham khảo nhé 

https://olm.vn/hoi-dap/detail/103171879928.html

\(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)

\(\Leftrightarrow\left(\frac{45-x}{1963}+1\right)+\left(\frac{40-x}{1968}+1\right)+\left(\frac{35-x}{1973}+1\right)+\left(\frac{30-x}{1978}+1\right)=0\)

\(\Leftrightarrow\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1973}=0\)

\(\Leftrightarrow\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

\(\Leftrightarrow x=2008\)

\(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)

=> \(\left(\frac{45-x}{1983}+1\right)+\left(\frac{40-x}{1968}+1\right)+\left(\frac{35-x}{1973}+1\right)+\left(\frac{30-x}{1978}+1\right)=0\)

=> \(\frac{2008-x}{1983}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=0\)

=> \(\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

Ta có \(\frac{1}{1963}>0,\frac{1}{1968}>0,\frac{1}{1973}>0,\frac{1}{1978}>0\)

=> \(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}>0\)

=> \(2008-x=0\)

=> \(x=2008\)