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

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\) 

\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)

Nên x + 1 = 0

=> x = -1

còn b vs c thì sao ạ

b) Sai đề à bạn đề \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)  hả đề này mk làm đc 

Bài làm

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

\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2+5}=\frac{35}{7}=5\)

Do đó: \(\hept{\begin{cases}\frac{x}{2}=5\\\frac{y}{-5}=5\end{cases}\Rightarrow\hept{\begin{cases}x=15\\y=-25\end{cases}}}\)

Vậy x = 15, y = -25

b) Ta thấy sai đề phần điều kiện x + ...

# Học tốt #

b, sửa lại là x+y =30

Ta có

\(2x=y4\Rightarrow\frac{x}{4}=\frac{y}{2}\)

Ap dụng tính chất DTSBN ta có

\(\frac{x}{4}=\frac{y}{2}=\frac{x+y}{4+20}=\frac{30}{6}=5\)

\(\hept{\begin{cases}\frac{x}{4}=5\\\frac{y}{2}=5\end{cases}\Rightarrow\hept{\begin{cases}x=20\\y=10\end{cases}}}\)

a \(\frac{x}{2}\)=\(\frac{y}{-5}\)và x-y=35

Áp dụng tính chất dãy tỉ số=nhau:\(\frac{x}{2}\)= \(\frac{y}{-5}\)=\(\frac{x-y}{2+5}\)=\(\frac{35}{7}\)=5

\(\hept{\begin{cases}x=10\\y=-25\end{cases}}\)

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

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

=> \(2008-x=0\)

=> \(x=2008\)