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

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

\(\Rightarrow\left(\dfrac{45-x}{1963}+1\right)+\left(\dfrac{40-x}{1968}+1\right)+\left(\dfrac{35-x}{1973}+1\right)+\left(\dfrac{30-x}{1978}+1\right)=-4+1+1+1+1\)

\(\Rightarrow\dfrac{2008-x}{1963}+\dfrac{2008-x}{1968}+\dfrac{2008-x}{1973}+\dfrac{2008-x}{1978}=0\)

Vì \(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\ne0\) nên 2008 - x = 0

\(\Rightarrow x=2008\)

\(\dfrac{55-x}{1963}\) + \(\dfrac{50-x}{1968}\) + \(\dfrac{45-x}{1973}\) + \(\dfrac{40-x}{1978}\) + 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, \(x=2018.\)

tham khảo đi Xem câu hỏi

bài này ra x=2013

Suy ra các số hang =-1 

45-x =-1968

suy ra x = 2013

Vậy x =2013

Các câu dễ tự làm :v

\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1981}=-4\) (sau khi đã sửa đề)

\(\Rightarrow\left(\dfrac{45-x}{1968}+1\right)+\left(\dfrac{40-x}{1973}+1\right)+\left(\dfrac{35-x}{1978}+1\right)+\left(\dfrac{30-x}{1981}+1\right)=0\)\(\Rightarrow\dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1981}=0\)

\(\Rightarrow\left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1981}\right)=0\)

\(\Rightarrow2013-x=0\Rightarrow x=2013\)

\(1+5+9+13+17+.....+x=5050\)

Số các số hạng là:

\(\dfrac{x-1}{4}+1=\dfrac{1}{4}x+\dfrac{3}{4}\)

Như vậy có :

\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\) số hạng

Theo đề bài ta có:

\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\left(x+1\right)=5050\)

\(\Rightarrow\left(\dfrac{1}{4}x+\dfrac{3}{4}\right)\left(x+1\right)=10100\)

\(\Rightarrow\dfrac{1}{4}x^2+\dfrac{1}{4}x+\dfrac{3}{4}x+\dfrac{3}{4}=10100\)

\(\Rightarrow\dfrac{1}{4}x^2+x+\dfrac{3}{4}=10100\)

Kiệt sức.đến đây ko nghĩ nổi nx

a,

\(5^x+5^{x+2}=650\\ 5^x\left(1+5^2\right)=650\\ 5^x\cdot26=650\\ 5^x=25\\ 5^x=5^2\\ \Rightarrow x=2\)

Vậy \(x=2\)

b,

\(\left(x+2\right)^2=81\\ \Rightarrow\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)

Vậy \(x=7\) hoặc \(x=-11\)

d,

\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}=-4\\ \dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}+4=0\\ \dfrac{45-x}{1968}+1+\dfrac{40-x}{1973}+1+\dfrac{35-x}{1978}+1+\dfrac{30-x}{1983}+1=0\\ \dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1983}=0\\ \left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\right)=0\)

Vì \(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\ne0\) nên

\(2013-x=0\\ x=2013\)

Vậy \(x=2013\)

e,

\(\dfrac{1}{2016}:2015x=\dfrac{-1}{2015}\\ 2015x=\dfrac{-2015}{2016}\\ x=\dfrac{-1}{2016}\)

Vậy \(x=\dfrac{-1}{2016}\)

Câu c mk bt lm nhưng sợ bị nói ăn cắp bản quyền của @Tuấn Anh Phan Nguyễn nên để bn ấy lm

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

Vì \(\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}+\frac{1}{1983}\ne0\)

\(\Rightarrow2013-x=0\)

\(\Rightarrow x=2013-0\)

\(\Rightarrow x=2013\)

Vậy \(x=2013.\)

Chúc bạn học tốt!

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