Question

[oops]

ghi kết quả thôi ạ , k phải trình bày ạ , e cần gấp ạ !!!!!

Answer 1

[Toru]

\(\dfrac{5}{8}+\dfrac{3}{8}x=1\dfrac{1}{8}\)

\(\Rightarrow\dfrac{5}{8}+\dfrac{3}{8}x=\dfrac{9}{8}\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{9}{8}-\dfrac{5}{8}\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{4}{8}\)

\(\Rightarrow x=\dfrac{4}{8}:\dfrac{3}{8}\)

\(\Rightarrow x=\dfrac{4}{8}\cdot\dfrac{8}{3}=\dfrac{4}{3}\)

\(---\)

\(\dfrac{1}{2}-\left(\dfrac{1}{3}x-2\right)=\dfrac{2}{3}\)

\(\Rightarrow\dfrac{1}{3}x-2=\dfrac{1}{2}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{1}{3}x-2=-\dfrac{1}{6}\)

\(\Rightarrow\dfrac{1}{3}x=-\dfrac{1}{6}+2\)

\(\Rightarrow\dfrac{1}{3}x=\dfrac{11}{6}\)

\(\Rightarrow x=\dfrac{11}{6}:\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{11}{6}\cdot3=\dfrac{11}{2}\)

\(---\)

\(\left(3x+4\right)\left(2x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x+4=0\\2x-3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=-4\\2x=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(--\)

\(-4+4x=9x-14\)

\(\Rightarrow4x-9x=-14+4\)

\(\Rightarrow-5x=-10\)

\(\Rightarrow x=\left(-10\right):\left(-5\right)=2\)

\(---\)

\(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)

\(=\left[\left(-\dfrac{14}{25}\right)^2\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}-\dfrac{1}{6}\)

\(=\left(\dfrac{196}{625}\cdot\dfrac{125}{49}\right)\cdot\dfrac{5}{6}-\dfrac{1}{6}\)

\(=\dfrac{4}{5}\cdot\dfrac{5}{6}-\dfrac{1}{6}\)

\(=\dfrac{4}{6}-\dfrac{1}{6}\)

\(=\dfrac{3}{6}\)

\(=\dfrac{1}{2}\)

\(Toru\)