Đáp án:
\(\begin{array}{l}
6)\quad B.\ 2\sqrt2a^3\\
7)\quad A.\ x - 7y - 4z + 9 =0\\
8)\quad D.\ I(1;-2;3),\ R = 4\\
9)\quad D.\ 7
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
6)\\
AB = \dfrac{AC}{\sqrt2}=\dfrac{2a}{\sqrt2} = a\sqrt2\\
\Rightarrow V = AB^3 = \left(a\sqrt2\right)^3 = 2a^3\sqrt2\\
7)\\
d: \dfrac{x-2}{1}=\dfrac{y-1}{-1}=\dfrac{z-1}{2}\\
có:\ \begin{cases}M(2;1;1)\in d\\VTCP\ \overrightarrow{u_d} = (1;-1;2)\end{cases}\\
\text{Ta có:}\\
d\subset (P) \Leftrightarrow \begin{cases}M(2;1;1)\in (P)\\VTCP\ \overrightarrow{u_P}=\overrightarrow{u_d} = (1;-1;2)\end{cases}\\
\text{Lại có:}\\
A(-2;1;0)\in (P)\\
\Rightarrow \overrightarrow{AM}=(-4;0;-1)\ \text{là VTCP của $(P)$}\\
\Rightarrow \overrightarrow{n} = \left[\overrightarrow{u_P};\overrightarrow{AM}\right]=(1;-7;-4)\ \text{là VTPT của $(P)$}\\
\text{Khi đó:}\\
(P): 1(x+2) - 7(y-1) - 4z =0\Leftrightarrow x - 7y - 4z +9 =0\\
8)\\
\quad (S):x^2 + y^2 + z^2 - 2x + 4y - 6z - 2 =0\\
\Leftrightarrow (x^2 - 2x + 1) + (y^2 + 4y + 4) + (z^2 - 6z + 9) = 16\\
\Leftrightarrow (x-1)^2 + (y+2)^2 + (z-3)^2 = 16\\
(S)\ \text{có tâm}\ I(1;-2;3),\ \text{bán kính}\ R = 4\\
9)\\
\quad \left(\dfrac15\right)^{\displaystyle{x-x^2}} = 5^{\displaystyle{6x-10}}\\
\Leftrightarrow 5^{\displaystyle{x^2 - x}}=5^{\displaystyle{6x-10}}\\
\Leftrightarrow x^2 - x = 6x - 10\\
\Leftrightarrow x^2 - 7x + 10 =0\\
\Rightarrow x_1 + x_2 = 7\quad (Theo\ Viète)
\end{array}\)
