Cho tam giác \[ABC\]có \[a = 8,b = 3,\widehat {C\,} = 120^\circ .\] Khi đó diện tích tam giác \[ABC\] bằng

Ta có: \(\left\{ \begin{array}{l}a,\,b,c > 0\\{a^3} = {b^3} + {c^3}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}0 < b < a\\0 < c < a\end{array} \right. \Rightarrow \left\{ \begin{array}{l}0 < \frac{b}{a} < 1\\0 < \frac{c}{a} < 1\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\left( {\frac{b}{a}} \right)^3} < {\left( {\frac{b}{a}} \right)^2}\\{\left( {\frac{c}{a}} \right)^3} < {\left( {\frac{c}{a}} \right)^2}\end{array} \right.\)

\( \Rightarrow {\left( {\frac{b}{a}} \right)^3} + {\left( {\frac{c}{a}} \right)^3} < {\left( {\frac{b}{a}} \right)^2} + {\left( {\frac{c}{a}} \right)^2} \Rightarrow \frac{{{b^3} + {c^3}}}{{{a^3}}} < \frac{{{b^2} + {c^2}}}{{{a^2}}}\)

\( \Rightarrow 1 < \frac{{{b^2} + {c^2}}}{{{a^2}}} \Rightarrow {b^2} + {c^2} - {a^2} > 0 \Rightarrow \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} > 0\)

Suy ra góc \(\widehat {BAC}\) nhọn.

\({a^3} = {b^3} + {c^3} = \left( {b + c} \right)\left( {{b^2} - bc + {c^2}} \right) > a\left( {{b^2} - bc + {c^2}} \right)\) \( \Rightarrow {a^2} > {b^2} - bc + {c^2} \Rightarrow \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} < \frac{1}{2} \Rightarrow \cos A < \cos {60^0}\)

Vậy \(\widehat {BAC} > {60^0}\).

a) Hàm số xác định khi \(\left\{ \begin{array}{l}5 - 2x \ge 0\\1 + 3x \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le \frac{5}{2}\\x \ge - \frac{1}{3}\end{array} \right.\).

Vậy tập xác định \(D = \left[ { - \frac{1}{3};\frac{5}{2}} \right]\).

b) Hàm số xác định khi \(\left\{ \begin{array}{l}x + 4 \ge 0\\{x^2} + 3x - 10 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge - 4\\x \ne - 5\\x \ne 2\end{array} \right.\).

Vậy tập xác định \(D = \left[ { - 4; + \infty } \right)\backslash \left\{ 2 \right\}\).