Question

Tìm giá trị của x để:

\(\left(18,6\cdot\frac{16\cdot x-3}{2}+\frac{139,5-238,25\cdot x}{5}\right):\left(1+8+15+22+...+282+288\right)=\frac{1}{3}\)

 

Answer 1

\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right):\left(1+8+15+...+281+288\right)=\frac{1}{3}\)

\(\Rightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right):\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)

\(\Rightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right):6069=\frac{1}{3}\)

\(\Rightarrow\frac{1488x-279+279-476,5x}{10}=2023\)

\(\Rightarrow1488x-476,5x=20230\)

\(\Rightarrow1011,5x=20230\)

\(\Rightarrow x=20\)

Answer 2

             Bài làm :

Ta có :

\(\left(18,6.\frac{16x-3}{2}+\frac{139,5-238,25x}{5}\right)\div\left(1+8+15+...+281+288\right)=\frac{1}{3}\)

\(\Leftrightarrow\left(\frac{297,6x-55,8}{2}+\frac{139,5-238,25x}{5}\right)\div\left(\frac{\left(288+1\right).\left[\left(288-1\right):\left(8-1\right)+1\right]}{2}\right)=\frac{1}{3}\)

\(\Leftrightarrow\left(\frac{1488x-279}{10}+\frac{279-476,5x}{10}\right)\div6069=\frac{1}{3}\)

\(\Leftrightarrow\frac{1488x-279+279-476,5x}{10}=2023\)

\(\Leftrightarrow1488x-476,5x=20230\)

\(\Leftrightarrow1011,5x=20230\)

\(\Leftrightarrow x=20\)

Vậy x=20