Ôn tập toán 6

NT

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

\(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)

\(60\%x=\dfrac{1}{3}.6\dfrac{1}{3}\)

\(\left(2x+\dfrac{3}{2}\right)^2-\dfrac{9}{25}=0\)

\(60\%x+0,4x+x:=2\)

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

\(-2x-\dfrac{-3}{5}:\left(-0,5\right)^2=-1\dfrac{1}{4}\)

NQ
2 tháng 5 2017 lúc 16:01

a) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\)

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

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

x=\(\dfrac{1}{10}:-\dfrac{2}{3}\)

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\).

b) \(\dfrac{1}{3}+\dfrac{2}{3}:x=-7\)

\(\dfrac{2}{3}:x=-7-\dfrac{1}{3}\)

\(\dfrac{2}{3}:x=-\dfrac{22}{3}\)

\(x=\dfrac{2}{3}:-\dfrac{22}{3}\)

\(x=-\dfrac{1}{11}\)

Vậy \(x=-\dfrac{1}{11}\).

c) \(60\%x=\dfrac{1}{3}\cdot6\dfrac{1}{3}\)

\(60\%x=\dfrac{19}{9}\)

\(\dfrac{3}{5}x=\dfrac{19}{9}\)

\(x=\dfrac{19}{9}:\dfrac{3}{5}\)

\(x=\dfrac{95}{27}\)

Vậy \(x=\dfrac{95}{27}\).

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

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

\(\dfrac{2}{3}-x=\dfrac{3}{20}\)

\(x=\dfrac{2}{3}-\dfrac{3}{20}\)

\(x=\dfrac{31}{60}\)

Vậy \(x=\dfrac{31}{60}\).

e) \(-2x-\dfrac{-3}{5}:\left(-0.5\right)^2=-1\dfrac{1}{4}\)

\(-2x-\dfrac{-12}{5}=-1\dfrac{1}{4}\)

\(-2x=-1\dfrac{1}{4}+\dfrac{-12}{5}\)

\(-2x=-\dfrac{73}{20}\)

\(x=-\dfrac{73}{20}:\left(-2\right)\)

\(x=\dfrac{73}{40}\)

Vậy \(x=\dfrac{73}{40}\).

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