Bài 3:
a) \(\dfrac{2x-1}{5}\)-\(\dfrac{x-2}{3}\)
=\(\dfrac{x+7}{15}\)
b) \(\dfrac{x+3}{2}\)-\(\dfrac{x-1}{3}\)
=\(\dfrac{x+5}{6}\)+1
c) \(\dfrac{2\left(x+5\right)}{3}\)+\(\dfrac{x+12}{2}\)
-\(\dfrac{5\left(x-2\right)}{6}\)=\(\dfrac{x}{3}\)+11
d) \(\dfrac{x-4}{5}\)+\(\dfrac{3x-2}{10}\)-x
=\(\dfrac{2x-5}{3}\)-\(\dfrac{7x+2}{6}\)
e) \(\dfrac{\left(2x-3\right)\left(2x+3\right)}{8}\)
=\(\dfrac{\left(x-4^{ }\right)^2}{6}\)+\(\dfrac{\left(x-2\right)^2}{3}\)
d) \(\dfrac{7x^2-14x-5}{15}\)
=\(\dfrac{\left(2x+1\right)^2}{5}\)-\(\dfrac{\left(x-1\right)^2}{3}\)
e) \(\dfrac{\left(7x+1\right)\left(x-2\right)}{10}\)+\(\dfrac{2}{5}\)
=\(\dfrac{\left(x-2\right)^2}{5}\)+\(\dfrac{\left(x-1\right)\left(x-3\right)}{2}\)
a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)
\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)
\(\Leftrightarrow6x-3-5x+10-x-7=0\)
\(\Leftrightarrow0x=0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
