Ca này bó tay rồi, ghi rõ ra đc ko, khó hiểu quá

\(x:0,25+86\cdot x+x:0,1=82,4\)

\(x\cdot4+86\cdot x+x\cdot10=82,4\)

\(x\left(4+86+10\right)=82,4\)

\(x\cdot100=82,4\\ x=82,4:100\\ x=0,824\)

CHẳng hỉu j

(x+1)+(x+2)+(x+3)+...+(x+100)=2020

x+1+x+2+x+3+...+x+100=2020

(x+x+x+...+x)+(1+2+3+...+100)=2020

100x+5050=2020

100x=2020-5050

100x=-3030

      x=-30,3

Ta có : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2009}{2011}\)

Đặt tổng vế trái là A

Ta có : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\)

\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)\div2}\right)\)

\(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}\)

\(\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\)

\(\frac{1}{2}A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{x}-\frac{1}{x+1}\right)\)

\(\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}\)

\(A=\left(\frac{1}{2}+\frac{1}{x+1}\right):\frac{1}{2}\)

\(A=1+\frac{1}{\left(x+1\right)\div2}\)

\(\Rightarrow1+\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}\)

\(\Rightarrow\frac{1}{\left(x+1\right)\div2}=\frac{2009}{2011}-1=\frac{2009}{2011}-\frac{2011}{2011}=-\frac{2}{2011}\)

\(\Rightarrow-\frac{2}{-\left(x+1\right)}=-\frac{2}{2011}\)

\(\Rightarrow-\left(x+1\right)=2011\)

\(\Rightarrow x+1=-2011\) 

\(\Rightarrow x=-2011-1=-2012\)

\(3^{x+2}+3^x=10\)

\(3^x.3^2+3^x.1=10\)

\(3^x\left(9+1\right)=10\)

\(3^x.10=10\)

\(3^x=1\)

\(\Rightarrow x=0\)

3^x+2+3^x=10

3^x+(2+1)=10

3^x+3=10

3^x=10-3=7

3^x=7

đề bài này sai xem lại nhé (vì 3^x không thể =7)

Đề bài bạn sai ý , bn đọc nhầm đề rồi , bn dưới mới đúng