Bài 6.17 trang 185 SBT đại số 10

LG a

\(A = {{2\tan \alpha - 3\cot \alpha } \over {\cos \alpha + tan\alpha }}\)

Lời giải chi tiết:

\(\displaystyle {\pi \over 2} < \alpha < \pi = > \cos \alpha < 0\)

Ta có: \(\displaystyle {\sin ^2}\alpha + {\cos ^2}\alpha = 1 \) \(\displaystyle \Rightarrow {\cos ^2}\alpha = 1 - {\sin ^2}\alpha \)

\(\displaystyle \Rightarrow \cos \alpha = - \sqrt {1 - {{\sin }^2}\alpha } \) \(\displaystyle = - \sqrt {1 - {9 \over {16}}} = - {{\sqrt 7 } \over 4}\)

\(\displaystyle \tan \alpha = {{\sin \alpha } \over {\cos \alpha }} = - {3 \over {\sqrt 7 }},\) \(\displaystyle \cot \alpha = - {{\sqrt 7 } \over 3}\)

Vậy \(\displaystyle A = \dfrac{{2.\left( { - \dfrac{3}{{\sqrt 7 }}} \right) - 3.\left( { - \dfrac{{\sqrt 7 }}{3}} \right)}}{{ - \dfrac{{\sqrt 7 }}{4} - \dfrac{3}{{\sqrt 7 }}}}\) \(\displaystyle = {{ - {6 \over {\sqrt 7 }} + \sqrt 7 } \over { - {{\sqrt 7 } \over 4} - {3 \over {\sqrt 7 }}}} = - {4 \over {19}}\)

LG b

\(B = {{{\rm{co}}{{\rm{s}}^2}\alpha + {{\cot }^2}\alpha } \over {\tan \alpha - \cot \alpha }}\)

Lời giải chi tiết:

\(\displaystyle B = \frac{{{{\left( { - \frac{{\sqrt 7 }}{4}} \right)}^2} + {{\left( { - \frac{{\sqrt 7 }}{3}} \right)}^2}}}{{ - \frac{3}{{\sqrt 7 }} - \left( { - \frac{{\sqrt 7 }}{3}} \right)}}\) \(\displaystyle = {{{7 \over {16}} + {7 \over 9}} \over { - {3 \over {\sqrt 7 }} + {{\sqrt 7 } \over {3 }}}} = {{{{7 \times 25} \over {144}}} \over { - {2 \over {3\sqrt 7 }}}} = - {{175\sqrt 7 } \over {96}}\)

xemloigiai.com