Nếu tam giác $ABC$  có \({a^2} < {b^2} + {c^2}\) thì

\(BC = 3 + 3\sqrt 6 .\)

\(BC = 3\sqrt 6  - 3.\)

\(BC = 3\sqrt 7 .\)

\(BC = \dfrac{{3 + 3\sqrt {33} }}{2}.\)

\(P \Rightarrow Q\)

\(Q \Rightarrow P\)

\(P \Leftrightarrow Q\)

\(P \Rightarrow P\)

$n-p$

$m + p$

$m-p$

$n + p$

$A \cup \emptyset  = A$

$A \cup A = A$

$\emptyset  \cup \emptyset  = \emptyset $

$\emptyset  \cup A = \emptyset $ .

\(y = 2{\left( {x + 3} \right)^2}\).

\(y = 2{x^2} + 3\)

\(y = 2{\left( {x - 3} \right)^2}\).

\(y = 2{x^2} - 3\).

\(y = \sqrt 2 {x^2} + 1\).

\(y =  - \sqrt 2 {x^2} + 1\).

\(y = \sqrt 2 {\left( {x + 1} \right)^2}\).

\(y = - \sqrt 2 {\left( {x + 1} \right)^2}\).

$x = 3.$

\(x = 3\) hoặc \(x = 0.\)

\(x =  \pm 3.\)

\(x =  \pm 1\).