b, 0,5 .\(\sqrt{100}\)-\(\sqrt{\frac{1}{4}}\)
=\(\frac{1}{2}\).10-\(\frac{1}{2}\)
=\(\frac{1}{2}\).(10-1)
=\(\frac{1}{2}\).9
=\(\frac{9}{2}\)
a) \(\frac{-7}{12}+\left(\frac{1}{4}+\frac{2}{3}\right)\)
\(=\frac{-7}{12}+\left(\frac{3}{12}+\frac{8}{12}\right)\)
\(=\frac{-7}{12}+\frac{11}{12}\)
\(=\frac{5}{12}\)
b) \(0,5.\sqrt{100}-\sqrt{\frac{1}{4}}\)
\(=\frac{1}{2}.10-\frac{1}{2}\)
\(=\frac{5}{1}-\frac{1}{2}\)
\(=\frac{10}{2}-\frac{1}{2}\)
\(=\frac{9}{2}\)
c,\(\left(-3\right)^2\).(\(\frac{3}{4}\)-0,25)-|\(3\frac{1}{2}\)-\(1\frac{1}{2}\)|
=9.(\(\frac{3}{4}\)-\(\frac{1}{4}\))-|\(\frac{7}{2}\)-\(\frac{3}{2}\)|
=9.\(\frac{2}{4}\)-\(\frac{4}{2}\)
=9.\(\frac{1}{2}\)-2
=\(\frac{9}{2}\)-2
=\(\frac{9}{2}\)-\(\frac{4}{2}\)
=\(\frac{5}{2}\)