Violympic toán 6

Phạm Thị Hà Giang

Giúp mình giải bài này vs:

\(\dfrac{30}{100}\).x+\(\dfrac{1}{4}\) = \(\dfrac{1}{5}\) .x-\(\dfrac{1}{2}\)

(\(\dfrac{1}{7.9}\) +\(\dfrac{1}{9.11}\) +........+\(\dfrac{1}{31.33}\)).x=(0,25-3,5).\(\dfrac{27}{3}\)

Thank you!

Soccer Kunkun
26 tháng 3 2017 lúc 19:12

1.

\(\dfrac{30}{100}.x+\dfrac{1}{4}=\dfrac{1}{5}.x-\dfrac{1}{2}\)

\(\dfrac{3}{10}.x=\dfrac{1}{5}.x-\dfrac{1}{2}-\dfrac{1}{4}\)

\(\dfrac{3}{10}.x=\dfrac{1}{5}.x-\dfrac{1}{4}\)

\(\dfrac{1}{5}.x-\dfrac{3}{10}.x=\dfrac{1}{4}\)

\(\left(\dfrac{1}{5}-\dfrac{3}{10}\right).x=\dfrac{1}{4}\)

\(\dfrac{-1}{10}.x=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}:\dfrac{-1}{10}\)

x=\(\dfrac{5}{-2}\)=\(\dfrac{-5}{2}\)

2.

\(\left(\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{31.33}\right).x=\left(0,25-3,5\right).\dfrac{27}{3}\)

\(\dfrac{2}{2}.\left(\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{31.33}\right).x=-3,25.9\)

\(\dfrac{1}{2}.\left(\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{31.33}\right).x=-29,25\)

\(\dfrac{1}{2}.\left(\dfrac{1}{7}-\dfrac{1}{33}\right).x=-29,25\)

\(\dfrac{1}{2}.\dfrac{26}{231}.x=-29,25\)

\(\dfrac{13}{231}.x=-29,25\)

\(x=-29,25:\dfrac{13}{231}\)

\(x=\dfrac{-2079}{4}\)

tick mink nha :)


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