H24
NL
3 tháng 7 2021 lúc 15:28

Bài 2 :

\(a,PT\Leftrightarrow\dfrac{11}{13}-\dfrac{5}{42}+x=-\dfrac{15}{28}+\dfrac{11}{13}\)

\(\Leftrightarrow x=-\dfrac{15}{28}+\dfrac{11}{13}-\dfrac{11}{13}+\dfrac{5}{42}=-\dfrac{5}{12}\)

Vậy ....

\(b,PT\Leftrightarrow x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{2}{5}\)

Vậy ...

\(c,PT\Leftrightarrow\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{5}{7}=-\dfrac{29}{70}\)

\(\Leftrightarrow x=-\dfrac{87}{140}\)

Vậy ...

\(d,PT\Leftrightarrow x=\dfrac{3}{7}-\dfrac{1}{4}+\left(-\dfrac{3}{5}\right)=-\dfrac{59}{140}\)

Vậy ...

\(e,PT\Leftrightarrow\dfrac{-21}{13}x=-\dfrac{2}{3}-\dfrac{1}{3}=-1\)

\(\Leftrightarrow x=\dfrac{13}{21}\)

Vậy ...

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NL
3 tháng 7 2021 lúc 15:43

Bài 3 :

a, \(PT\Leftrightarrow\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=-\dfrac{3}{20}\)

Vậy ...

b,\(PT\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

Vậy ...

c, \(PT\Leftrightarrow\dfrac{1}{4x}=\dfrac{2}{5}-\dfrac{3}{4}=-\dfrac{7}{20}\)

\(\Leftrightarrow4x=-\dfrac{20}{7}\)

\(\Leftrightarrow x=-\dfrac{5}{7}\)

Vậy .....

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