Question

a/\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)            

b/ \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)

Answer 1

a)

\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

= \(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

= \(\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)

= \(\frac{1}{3}+\left(-1\right)+1\)

= \(\frac{1}{3}+0=\frac{1}{3}\)

b)

\(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)

= \(\frac{38}{45}-\frac{8}{45}+\frac{17}{51}+\frac{3}{11}\)

= \(\left(\frac{38}{45}-\frac{8}{45}\right)+\left(\frac{17}{51}+\frac{3}{11}\right)\)

= \(\frac{2}{3}+\frac{20}{33}\)

= \(\frac{14}{11}\)