Đáp án:
$B = \dfrac{{71}}{{125}}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
\sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{5}\\
\Leftrightarrow \sin x\cos \dfrac{\pi }{4} + \sin \dfrac{\pi }{4}\cos x = \dfrac{{\sqrt 2 }}{5}\\
\Leftrightarrow \dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = \dfrac{{\sqrt 2 }}{5}\\
\Leftrightarrow \sin x + \cos x = \dfrac{2}{5}\\
\Rightarrow {\left( {\sin x + \cos x} \right)^2} = \dfrac{4}{{25}}\\
\Leftrightarrow {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x = \dfrac{4}{{25}}\\
\Leftrightarrow 1 + 2\sin x\cos x = \dfrac{4}{{25}}\\
\Leftrightarrow \sin x\cos x = \dfrac{{ - 21}}{{50}}
\end{array}$
Khi đó:
$\begin{array}{l}
B = {\sin ^3}x + {\cos ^3}x\\
= {\left( {\sin x + \cos x} \right)^3} - 3\sin x\cos x\left( {\sin x + \cos x} \right)\\
= {\left( {\dfrac{2}{5}} \right)^3} - 3.\left( {\dfrac{{ - 21}}{{50}}} \right).\dfrac{2}{5}\\
= \dfrac{{71}}{{125}}
\end{array}$
Vậy $B = \dfrac{{71}}{{125}}$
