Cho hình vẽ, biết \(\widehat {aBE} = 60^\circ \), \(\widehat {BED} = 60^\circ \); \(\widehat {BCD} = 135^\circ \).

Hỏi số đo \(\widehat {CDE}\) bằng bao nhiêu độ?

Hướng dẫn giải

a) \(\frac{{13}}{{25}} - \frac{{11}}{4} - \frac{{38}}{{25}} + \frac{{15}}{4} + \frac{1}{2} = \left( {\frac{{13}}{{25}} - \frac{{28}}{{25}}} \right) + \left( {\frac{{15}}{4} - \frac{{11}}{4}} \right) + \frac{1}{2} = \frac{{ - 15}}{{25}} + 1 + \frac{1}{2} = \frac{{ - 3}}{5} + 1 + \frac{1}{2} = \frac{2}{5} + \frac{1}{2} = \frac{9}{{10}}\).

b) \({\left( {\frac{1}{2} - \frac{2}{3}} \right)^2} + 1\frac{2}{3}:\left| { - 0,75} \right| - \sqrt {\frac{1}{{16}}} = {\left( { - \frac{1}{6}} \right)^2} + \frac{5}{3}:0,75 - \frac{1}{4}\)

                                                     \( = \frac{1}{{36}} + \frac{5}{3}.\frac{4}{3} - \frac{1}{4}\)

                                                     \( = \frac{1}{{36}} + \frac{{20}}{9} - \frac{1}{4}\)

                                                    \( = \frac{1}{{36}} + \frac{{80}}{{36}} - \frac{9}{{36}} = \frac{{72}}{{36}} = 2\).

a) Ta có \(\widehat {xOz}\) và \(\widehat {zOy}\) là hai góc kề nhau nên \(\widehat {xOz} + \widehat {zOy} = \widehat {xOy}\), mà \(\widehat {xOy}\) là góc bẹt nên \(\widehat {xOy} = 180^\circ \)

Do đó, \(\widehat {xOz} + \widehat {zOy} = 180^\circ \), suy ra \(\widehat {zOy} = 180^\circ - \widehat {xOz} = 180^\circ - 70^\circ = 110^\circ \).

Vậy \(\widehat {zOy} = 110^\circ \).