Question

[Lê Phương Uyên]

\(\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{19.21}\right).462-\left[0,04:\left(x+1,05\right)\right]:0,12=19\) hãy tìm x

Answer 1

[Đức Minh]

\(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)\cdot462-\left[0,04:\left(x+1,05\right)\right]:0,12=19\)\(\Leftrightarrow\dfrac{2}{2}\cdot\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\cdot462-\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=19\)\(\Leftrightarrow\left(\dfrac{1}{11}-\dfrac{1}{21}\right)\cdot462-\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=19\)

\(\Leftrightarrow20-\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=19\)

\(\Leftrightarrow\left[\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)\right]:\dfrac{3}{25}=1\)

\(\rightarrow\dfrac{1}{25}:\left(x+\dfrac{21}{20}\right)=\dfrac{3}{25}\)

\(\Leftrightarrow x+\dfrac{21}{20}=\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{1}{3}-\dfrac{21}{20}=-\dfrac{43}{60}\)

Vậy \(x=-\dfrac{43}{60}\)