Lớp nghìn gồm các chữ số: 4 (hàng nghìn), 9 (hàng chục nghìn), 6 (hàng trăm nghìn)

Ta có: 14 : 4 = 3 (dư 2). Vậy cần ít nhất 3 + 1 = 4 cái lồng. Đáp án B

\(-5x^4+18x^2-13=0\\ \Rightarrow\left(-5x^4+5x^2\right)+\left(13x^2-13\right)=0\\ \Rightarrow-5x^2\left(x^2-1\right)+13\left(x^2-1\right)=0\\ \Rightarrow\left(x^2-1\right)\left(-5x^2+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2-1=0\\-5x^2+13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2=1\\x^2=\dfrac{13}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm1\\x=\pm\sqrt{\dfrac{13}{5}}\end{matrix}\right.\)

\(\dfrac{2}{3}\times\dfrac{4}{5}-\dfrac{1}{3}\times\dfrac{4}{5}=\left(\dfrac{2}{3}-\dfrac{1}{3}\right)\times\dfrac{4}{5}=\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{15}\)

\(\dfrac{1}{2}:\dfrac{3}{4}+\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}+\dfrac{1}{6}\times\dfrac{4}{3}=\left(\dfrac{1}{2}+\dfrac{1}{6}\right)\times\dfrac{4}{3}=\dfrac{2}{3}\times\dfrac{4}{3}=\dfrac{8}{9}\)

\(\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{1}{6}:\dfrac{3}{4}=\dfrac{1}{2}\times\dfrac{4}{3}-\dfrac{1}{6}\times\dfrac{4}{3}=\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\times\dfrac{4}{3}=\dfrac{1}{3}\times\dfrac{4}{3}=\dfrac{4}{9}\)

\(m_a=\sqrt{\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}}=\sqrt{\dfrac{8^2+10^2}{2}-\dfrac{7^2}{4}}=\sqrt{\dfrac{279}{4}}=\dfrac{3\sqrt{31}}{2}\)