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a) đề kiểu gì vậy bạn
b) \(x+\frac{2}{14}=\frac{3}{21}\Leftrightarrow x=0\)
c) \(x+\frac{2}{7}=\frac{1}{2}\Leftrightarrow x=\frac{3}{14}\)
d) \(\frac{12}{15}=\frac{x-5}{10}\Leftrightarrow15x-75=120\Leftrightarrow15x=195\Leftrightarrow x=13\)
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
=> \(\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)
=> \(\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)
=> \(\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy ...
\(x-\frac{1}{15}=\frac{1}{10}\)
\(x=\frac{1}{10}+\frac{1}{15}\)
\(x=\frac{5}{3}\)
Vậy \(x=\frac{5}{3}\)
\(-\frac{2}{15}-x=-\frac{3}{10}\)
\(x=-\frac{2}{15}+\frac{3}{10}\)
\(x=\frac{1}{15}\)
Vậy \(x=\frac{1}{15}\)
\(x-\frac{1}{15}=\frac{1}{10}\)
\(\Rightarrow x=\frac{1}{10}+\frac{1}{15}\)
\(\Rightarrow x=\frac{1}{6}\)
\(\frac{-2}{15}-x=\frac{-3}{10}\)
\(\Rightarrow x=\frac{-2}{15}-\frac{-3}{10}\)
\(\Rightarrow x=\frac{1}{6}\)
x=-2/15+3/10
x=5/30
x=1/6 nha
-2/15-x=-3/10
x=1/6
HT