Question

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Answer 1

[꧁_͏ͥ͏_͏ͣ͏_͏ͫ͏ ¡亗ᵛᶰシ๖ۣۜT๖ۣۜR๖ۣۜÂ๖ۣ...]

Bài 1:

a) \(3\dfrac{4}{5}.\left(-1\dfrac{1}{2}\right)+3\dfrac{4}{5}.2\dfrac{1}{2}-1\dfrac{4}{7}\) 

\(=\dfrac{19}{5}.\left(\dfrac{-3}{2}+\dfrac{5}{2}\right)-\dfrac{11}{7}\) 

\(=\dfrac{19}{5}.1-\dfrac{11}{7}\) 

 \(=\dfrac{19}{5}-\dfrac{11}{7}\) 

\(=\dfrac{78}{35}\) 

b) \(\dfrac{3}{4}:\left(-2\dfrac{1}{2}\right)-\dfrac{4}{5}:\left(-2\dfrac{1}{2}\right)+1\dfrac{1}{4}:\left(-2\dfrac{1}{2}\right)\) 

\(=\left(\dfrac{3}{4}-\dfrac{4}{5}+\dfrac{5}{4}\right):\dfrac{-5}{2}\) 

\(=\dfrac{6}{5}:\dfrac{-5}{2}\) 

\(=\dfrac{-12}{25}\) 

c) \(\dfrac{4}{5}-\left[\dfrac{3}{5}.\left(\dfrac{-2}{3}-1\dfrac{1}{3}\right)+\dfrac{4}{7}\right]\) 

\(=\dfrac{4}{5}-\left[\dfrac{3}{5}.\left(\dfrac{-2}{3}-\dfrac{4}{3}\right)+\dfrac{4}{7}\right]\) 

\(=\dfrac{4}{5}-\left[\dfrac{3}{5}.-2+\dfrac{4}{7}\right]\) 

\(=\dfrac{4}{5}-\left[\dfrac{-6}{5}+\dfrac{4}{7}\right]\) 

\(=\dfrac{4}{5}-\left(\dfrac{-22}{35}\right)\) 

\(=\dfrac{10}{7}\)

Answer 2

[꧁_͏ͥ͏_͏ͣ͏_͏ͫ͏ ¡亗ᵛᶰシ๖ۣۜT๖ۣۜR๖ۣۜÂ๖ۣ...]

Bài 2:

a) \(\left(\dfrac{3}{4}-x\right).\dfrac{1}{2}-\dfrac{4}{5}=1\dfrac{1}{5}\) 

            \(\left(\dfrac{3}{4}-x\right).\dfrac{1}{2}=\dfrac{6}{5}+\dfrac{4}{5}\) 

             \(\left(\dfrac{3}{4}-x\right).\dfrac{1}{2}=2\) 

                       \(\dfrac{3}{4}-x=2:\dfrac{1}{2}\) 

                       \(\dfrac{3}{4}-x=4\) 

                              \(x=\dfrac{3}{4}-4\) 

                              \(x=\dfrac{-13}{4}\) 

b) \(\dfrac{3}{4}-\dfrac{1}{4}.x=2\dfrac{1}{3}\) 

            \(\dfrac{1}{4}.x=\dfrac{3}{4}-\dfrac{7}{3}\) 

            \(\dfrac{1}{4}.x=\dfrac{-19}{12}\) 

                 \(x=\dfrac{-19}{12}:\dfrac{1}{4}\) 

                 \(x=\dfrac{-19}{3}\)