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

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

(1 + \(\dfrac{55-x}{1963}\) ) + (  1 + \(\dfrac{50-x}{1968}\)) + (1+ \(\dfrac{45-x}{1973}\))+ (1 + \(\dfrac{40-x}{1978}\)) = 0

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

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

(2018 - \(x\))\(\times\)( \(\dfrac{1}{1963}\)+\(\dfrac{1}{1986}\)+\(\dfrac{1}{1973}\)+) =0

                              2018 \(-x\) = 0

                              \(x\) = 2018

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

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

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

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

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

Mà \(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\ne0\)

\(\Rightarrow\text{ }2018-x=0\)

\(\Rightarrow\text{ }x=2018-0\)

\(\Rightarrow\text{ }x=2018\)

Vậy, \(x=2018.\)

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

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

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

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

\(\Leftrightarrow x=2008\)

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

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

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

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

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

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

=> \(2008-x=0\)

=> \(x=2008\)

45-x/1963+40-x/1968+35-x/1973+30-x/1978+4=0

45-x/1963+40-x/1968+35-x/1973+30-x/1978=-4

(45-x/1963+1)+(40-x/1968+1)+(35-x/1973+1)+(30-x/1978+1)=-4+1+1+1+1

2008-x/1963+2008-x/1968+2008-x/1973+2008-x/1978=0

(2008-x).(1/1963+1/1968+1/1973+1/1978)=0

Vì 1/1963+1/1968+1/1973+1/1978 khac o

2008-x=0

x=2008

k cho minh dau tien nha!

anh em ơi hãy k cho tôi

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