\(3\dfrac{1}{3x}-6\dfrac{3}{4}=3\dfrac{1}{4}\)
\(\Rightarrow\dfrac{9x+1}{3x}-\dfrac{28}{4}=\dfrac{13}{4}\)
\(\Rightarrow\dfrac{9x+1}{3x}=\dfrac{41}{4}\)
\(\Rightarrow\left(9x+1\right)\cdot4=41\cdot3x\)
\(\Rightarrow x=\dfrac{4}{87}\)
\(3\dfrac{1}{3x}-6\dfrac{3}{4}=3\dfrac{1}{4}\)
\(\Rightarrow\dfrac{9x+1}{3x}-\dfrac{27}{4}=\dfrac{13}{4}\)
\(\Rightarrow4\left(9x+1\right)-81x=39x\)
\(\Rightarrow36x+4-81x-39x=0\Rightarrow84x=4\)
\(\Rightarrow x=\dfrac{4}{84}=\dfrac{1}{21}\)
ĐKXĐ: \(x\ne0\)
Ta có: \(3\cdot\dfrac{1}{3x}-6\dfrac{3}{4}=3\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{3x}=\dfrac{13}{4}+\dfrac{27}{4}\)
\(\Leftrightarrow\dfrac{1}{x}=10\)
hay \(x=\dfrac{1}{10}\)(thỏa ĐK)
Vậy: \(x=\dfrac{1}{10}\)
x=1
thay vào biểu thức:
\(3\dfrac{1}{3.3}-6\dfrac{3}{4}=3\dfrac{1}{4}\)
=>10- \(\dfrac{27}{4}=\dfrac{13}{4}\)
