Bài 104131

Question

$a)$ Chứng minh rằng:
$1/ tan5^0.tan55^0 tan65^0 = 2 - \sqrt 3 $
$2/ cos^2x + cos^2 \left( {\frac{{\pi }}{{3}}{ - x}} \right){ -
cosx}{.cos}\left( {\frac{{\pi }}{{3}}{ - x}} \right){ =
}\frac{{3}}{{4}}$
$b)$ Giải hệ phương trình: $\left\{ \begin{array}{l}
{sinx}{.siny = }\frac{{1}}{{4}}\\
{3tanx = coty}
\end{array} \right.$

score 0 · Answer 1

$a) 1/ tan5^0.tan55^0 tan65^0 = 2 - \sqrt 3 $
$VT = tan5^0.tan55^0 tan65^0 = tan5^0.tan(60^0 + 5^0)tan(60^0 + 5^0)$
       $\begin{array}{l}
{ = tan}{{5}^{0}}\frac{{{tan6}{{0}^{0}}{ -
tan}{{5}^{0}}}}{{{1 +
tan6}{{0}^{0}}{.tan}{{5}^{0}}}}{.}\frac{{{tan6}{{0}^{0}}{ + tan}{{5}^{0}}}}{{{1 -
tan6}{{0}^{0}}{.tan}{{5}^{0}}}}\\
{ = tan}{{5}^{0}}\frac{{\sqrt {3} { -
tan}{{5}^{0}}}}{{{1 + }\sqrt {3}
{tan}{{5}^{0}}}}{.}\frac{{\sqrt {3} { +
tan}{{5}^{0}}}}{{{1 - }\sqrt {3} {tan}{{5}^{0}}}}\\
{ = }\frac{{{tan}{{5}^{0}}{(3 -
t}{{an}^{2}}{{5}^{0}}{)}}}{{{1 -
3t}{{an}^{2}}{{5}^{0}}}}{ =
}\frac{{{3tan}{{5}^{0}}{ -
t}{{an}^{3}}{{5}^{0}}}}{{{1 -
3t}{{an}^{2}}{{5}^{0}}}}{ =
tan(3}{.}{{5}^{0}}{) = tan1}{{5}^{0}}\\
\left( {{tan3\alpha = }\frac{{{3tan\alpha - t}{{an}^{3}}{\alpha
}}}{{{1 - 3t}{{an}^{2}}{\alpha }}}} \right)\\
{ = tan(6}{{0}^{0}}{ - 45}{}^0{) =
}\frac{{{tan6}{{0}^{0}}{ - tan45}{}^0}}{{{1 +
tan6}{{0}^{0}}{.tan45}{}^0}}
\end{array}$
       = $\frac{{\sqrt {3} { - 1}}}{{{1 + }\sqrt {3} }}{ =
}\frac{{{{\left( {\sqrt {3} { - 1}} \right)}^{2}}}}{{{3 - 1}}}{ =
}\frac{{{4 - 2}\sqrt {3} }}{{2}}{ = 2 - }\sqrt {3} { = VP}$
   (đpcm)
$2/VT = cos^2x + cos^2 (\left( {\frac{{\pi }}{{3}}{ - x}} \right){ -
cosx}{.cos}\left( {\frac{{\pi }}{{3}}{ - x}} \right)$
         $\begin{array}{l}
{ = }{\left[ {{2sin}\frac{{\pi }}{{6}}{sin}\left( {\frac{{\pi
}}{{6}}{ - x}} \right)} \right]^{2}}{ +
}\frac{{1}}{{2}}\left[ {{cos}\frac{{\pi }}{{3}}{ + cos}\left(
{\frac{{\pi }}{{3}}{ - 2x}} \right)} \right]\\
{ = 2}{.}\frac{{1}}{{2}}{si}{{n}^2}\left( {\frac{{\pi
}}{{6}}{ - x}} \right){ +
}\frac{{1}}{{2}}{.}\frac{{1}}{{2}}{ +
}\frac{{1}}{{2}}\left[ {{1 - 2si}{{n}^2}\left( {\frac{{\pi
}}{{6}}{ - x}} \right)} \right]
\end{array}$
         ${ = }\frac{{1}}{{4}}{ +
}\frac{{1}}{{2}}{ = }\frac{{3}}{{4}}{ = VP}$ (đpcm)
$b$) Ta có: $3tanx=coty$
Điều kiện: $x \neq \frac{{\pi }}{{2}}{ + k2\pi , y \neq }{{k}^{,}}{\pi ,k, }{{k}^{,}} \in {Z}$
$ \Leftrightarrow \frac{{{sinx}}}{{{cosx}}}{ =
}\frac{{{cosy}}}{{{siny}}} \Leftrightarrow {3sinx}{.siny =
cosx}{.cosy}$
$ \Leftrightarrow {cosx}{.cosy = }\frac{{3}}{{4}}$ (vì $sinx.siny =\frac{1}{4}$)
Hệ: $ \Leftrightarrow \left\{ \begin{array}{l}
{sinx}{.siny = }\frac{{1}}{{4}}\\
{cosx}{.cosy = }\frac{{3}}{{4}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{cosx}{.cosy + sinx}{.siny = 1}\\
{cosx}{.cosy - sinx}{.siny = }\frac{{1}}{{2}}
\end{array} \right.$
        $\begin{array}{l}
\\
\Leftrightarrow \left\{ \begin{array}{l}
{cos(x - y) = 1}\\
{cos(x + y) = }\frac{{1}}{{2}}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x - y = k2\pi }\\
{x + y = \pm }\frac{{\pi }}{{3}}{ + l2\pi }
\end{array} \right.{    (k, l} \in {Z)}\\
\Leftrightarrow \left\{ \begin{array}{l}
{x = }\frac{{\pi }}{{6}}{ + (k + l)\pi }\\
{y = }\frac{{\pi }}{{6}}{ + ( l - k)\pi }
\end{array} \right.{ } \vee { }\left\{ \begin{array}{l}
{x = - }\frac{{\pi }}{{6}}{ + (l + k)\pi }\\
{y = - }\frac{{\pi }}{{6}}{ + (l - k)\pi }
\end{array} \right.
\end{array}$

Bài 104131

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Bài 104131

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