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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
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 ...
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
\(\frac{x+1}{1974}+1+\frac{x+2}{1973}+1+\frac{x+3}{1972}+1=-3+3\)
\(\Leftrightarrow\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)
\(\Leftrightarrow\left(x+1975\right)\left(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\right)=0\)
=> x + 1975 = 0. => x = -1975 ( vì 1/1974 + 1/1973 + 1/1972 khác 0 )