a) Cho bất phương trình ({x^2} - m(x - 1) ge 0)Tìm m để bất phương trình trên đúng với (forall x in R)b) Cho (c{

a) Đặt $f(x) = {x^2} - m(x - 1) = {x^2} - mx + m$. ycbt $f(x) \ge 0$ với mọi $x \in R$

- Ycbt $\Leftrightarrow \Delta  = {m^2} - 4m \le 0$

$\Leftrightarrow 0 \le m \le 4$

b) Ta có: ${\sin ^2}\alpha  = 1 - c{\rm{o}}{{\rm{s}}^2}\alpha  = 1 - \frac{{16}}{{25}} = \frac{9}{{25}} \Rightarrow \sin \alpha  =  \pm \frac{3}{5}$

Vì $\frac{\pi }{2} < \alpha  < \pi$ suy ra $\sin \alpha  > 0$ nên $\sin \alpha  = \frac{3}{5}$

$\begin{array}{l} A = \sin \left( {\frac{\pi }{4} - \alpha } \right) + c{\rm{os}}\left( {\alpha  + \frac{{5\pi }}{6}} \right) - \frac{{2\sqrt 3 }}{5}\\  = \sin \alpha .\cos \frac{\pi }{4} - c{\rm{os}}\alpha .\sin \frac{\pi }{4} + c{\rm{os}}\alpha .\cos \frac{{5\pi }}{6} - \sin \alpha .\sin \frac{{5\pi }}{6} - \frac{{2\sqrt 3 }}{5}\\  = \frac{3}{5}.\frac{{\sqrt 2 }}{2} + \frac{4}{5}.\frac{{\sqrt 2 }}{2} + \frac{4}{5}.\frac{{\sqrt 3 }}{2} - \frac{3}{5}.\frac{1}{2} - \frac{{2\sqrt 3 }}{5} = \frac{{ - 3 + 7\sqrt 2 }}{{10}} \end{array}$

c) Ta có: 

$\begin{array}{l} P = c{\rm{o}}{{\rm{s}}^2}\left( {\frac{\pi }{2} - x} \right) + c{\rm{o}}{{\rm{s}}^2}\left( {\pi  - x} \right) - 1 + \tan (\pi  + x).\cot (3\pi  - x)\\  = {\rm{co}}{{\rm{s}}^2}x + {\sin ^2}x - 1 - \tan x\cot x\\  =  - 1 \end{array}$