(1,5 điểm). Cho góc bẹt \(\widehat {xOy}\). Vẽ tia \(Oz\) sao cho \(\widehat {xOz} = 70^\circ \).

a) Tính \(\widehat {zOy}\).

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

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

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

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

Hướng dẫn giải

Đáp án: \( - 0,6\)

Ta có: \(\frac{2}{3}x + \frac{7}{{10}} = \frac{3}{{10}}\)

           \(\frac{2}{3}x = \frac{3}{{10}} - \frac{7}{{10}}\)

           \(\frac{2}{3}x = - \frac{2}{5}\)

           \(x = - \frac{2}{5}:\frac{2}{3}\)

           \(x = - \frac{2}{5}.\frac{3}{2}\)

           \(x = - \frac{3}{5}\) hay \(x = - 0,6.\)

Vậy \(x = - 0,6.\)