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