\(x\dfrac{x}{15}=\dfrac{112}{5}\)
\(\dfrac{x.15+x}{15}\)=\(\dfrac{112}{5}\)
\(\dfrac{x\left(15+1\right)}{15}=\dfrac{112}{5}\)
\(\dfrac{x.16}{15}=\dfrac{112}{5}\)
=> \(x.16.5=15.112\)
\(x.80=1680\)
=> \(x=21\)
\(x\dfrac{x}{15}=\dfrac{112}{5}\)
\(\dfrac{x.15+x}{15}=\dfrac{112}{5}\)
\(\dfrac{x\left(15+1\right)}{15}=\dfrac{112.3}{5.3}\)
\(\dfrac{x.16}{15}=\dfrac{366}{15}\)
\(\Rightarrow x.16=366\)
\(\Rightarrow x=21\)
\(x\dfrac{x}{15}\)=\(\dfrac{112}{5}\)
\(\dfrac{x.15+x}{15}=\dfrac{336}{15}\)
\(\dfrac{16x}{15}=\dfrac{336}{15}\)
⇒16x=336
x=336/16
x=21
Ta có:
\(x\dfrac{x}{15}=\dfrac{112}{5}\)
⇒ \(\dfrac{x.15+x}{15}=\dfrac{112}{5}\)
⇒ \(\dfrac{16x}{15}=\dfrac{336}{15}\)
⇒ 16x= 336
⇒ x= 21
\(x\dfrac{x}{15}=\dfrac{112}{5}\)
\(\Rightarrow\dfrac{x\cdot15+x}{15}=\dfrac{112}{15}\)
\(\Rightarrow\dfrac{x\left(15+1\right)}{15}=\dfrac{112}{15}\)
\(\Rightarrow\dfrac{x\cdot16}{15}=\dfrac{112}{15}\)
\(\Rightarrow16x=112\)
\(\Rightarrow x=112:16\)
\(\Rightarrow x=7\)
Vậy x=7
\(x\dfrac{x}{5}=\dfrac{112}{5}\)
\(\Rightarrow\dfrac{16x}{15}=\dfrac{112}{15}\)
\(\Rightarrow16x=112\)
\(\Rightarrow x=112:16\)
\(\Rightarrow x=7\)
Vậy x=7