Bài 2:
a: \(\dfrac{1}{3}x-\dfrac{2}{5}=-\dfrac{7}{15}\)
=>\(\dfrac{1}{3}x=-\dfrac{7}{15}+\dfrac{2}{5}=\dfrac{-7+6}{15}=\dfrac{-1}{15}\)
=>\(x=\dfrac{-1}{15}:\dfrac{1}{3}=\dfrac{-1}{15}\cdot3=-\dfrac{1}{5}\)
b: \(-2,5-x=\dfrac{15}{6}\)
=>-2,5-x=2,5
=>x=-2,5-2,5=-5
c: \(\left(\dfrac{2}{7}-x\right)^2=\dfrac{16}{49}\)
=>\(\left(x-\dfrac{2}{7}\right)^2=\dfrac{16}{49}\)
=>\(\left[{}\begin{matrix}x-\dfrac{2}{7}=\dfrac{4}{7}\\x-\dfrac{2}{7}=-\dfrac{4}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{7}\\x=-\dfrac{2}{7}\end{matrix}\right.\)
Bài 1:
a: \(\dfrac{13}{25}-\dfrac{31}{41}+\dfrac{12}{25}-\dfrac{10}{41}-0.5\)
\(=\left(\dfrac{13}{25}+\dfrac{12}{25}\right)-\left(\dfrac{31}{41}+\dfrac{10}{41}\right)-\dfrac{1}{2}\)
=1-1-0,5
=-0,5
b: \(\left(-2\right)^3-\left(-\dfrac{1}{2}\right)^2:\dfrac{-1}{16}-2023^0\)
\(=-8-1-\dfrac{1}{4}\cdot\left(-16\right)\)
=-9+4
=-5
c: \(\left(-\dfrac{1}{3}\right)^2+\dfrac{4}{3}:2-0,6\)
\(=\dfrac{1}{9}+\dfrac{2}{3}-\dfrac{3}{5}\)
\(=\dfrac{10+60-54}{90}=\dfrac{16}{90}=\dfrac{8}{45}\)
d: \(\dfrac{2}{9}\cdot\dfrac{7}{5}+\dfrac{2}{9}\cdot\dfrac{-11}{5}+\dfrac{4}{5}\cdot\dfrac{2}{9}\)
\(=\dfrac{2}{9}\left(\dfrac{7}{5}-\dfrac{11}{5}+\dfrac{4}{5}\right)\)
\(=\dfrac{2}{9}\cdot0=0\)