Đáp án:
B1:
1)$S = \left\{ { - 2;2} \right\}$
Giải thích các bước giải:
B1:
1) Ta có:
$\begin{array}{l}
{x^2} + 3x + 5 = \left( {x + 3} \right)\sqrt {{x^2} + 5} \\
\Leftrightarrow 2\left( {{x^2} + 3x + 5} \right) - 2\left( {x + 3} \right)\sqrt {{x^2} + 5} = 0\\
\Leftrightarrow \left( {{x^2} + 5} \right) - 2\sqrt {{x^2} + 5} .\left( {x + 3} \right) + {x^2} + 6x + 5 = 0\\
\Leftrightarrow \left( {{x^2} + 5} \right) - 2\sqrt {{x^2} + 5} .\left( {x + 3} \right) + {\left( {x + 3} \right)^2} - 4 = 0\\
\Leftrightarrow {\left( {\sqrt {{x^2} + 5} - x - 3} \right)^2} - 4 = 0\\
\Leftrightarrow \left( {\sqrt {{x^2} + 5} - x - 5} \right)\left( {\sqrt {{x^2} + 5} - x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {{x^2} + 5} = x + 5\\
\sqrt {{x^2} + 5} = x + 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 5 \ge 0\\
{x^2} + 5 = {\left( {x + 5} \right)^2}
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 \ge 0\\
{x^2} + 5 = {\left( {x + 1} \right)^2}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - 5\\
10x + 20 = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge - 1\\
2x - 4 = 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - 5\\
x = - 2
\end{array} \right.\left( c \right)\\
\left\{ \begin{array}{l}
x \ge - 1\\
x = 2
\end{array} \right.\left( c \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 2\\
x = 2
\end{array} \right.
\end{array}$
Vậy tập nghiệm của phương trình là: $S = \left\{ { - 2;2} \right\}$
2) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
a + b - 2c = 0\\
2ab - bc - ac = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a - c = c - b\\
b\left( {a - c} \right) + a\left( {b - c} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a - c = c - b\\
b\left( {c - b} \right) + a\left( {b - c} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a - c = c - b\\
\left( {c - b} \right)\left( {b - a} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a - c = c - b\left( 1 \right)\\
\left[ \begin{array}{l}
c = b\\
b = a
\end{array} \right.
\end{array} \right.\\
+ )TH1:c = b\\
\left( 1 \right) \Leftrightarrow a - c = c - b = 0\\
\Rightarrow a = c = b\\
+ )TH2:b = a\\
\left( 1 \right) \Leftrightarrow a - c = c - a\\
\Leftrightarrow c = a\\
\Rightarrow a = b = c
\end{array}$
Vậy ta có điều phải chứng minh.
B2:
Ta có:
$A = {11^n} + {7^n} - {2^n} - 1$
+) Lại có:
$\begin{array}{l}
11 \equiv 2\left( {\bmod 3} \right)\\
\Rightarrow {11^n} \equiv {2^n}\left( {\bmod 3} \right)\\
\Rightarrow {11^n} - {2^n} \equiv 0\left( {\bmod 3} \right)\left( 1 \right)
\end{array}$
Và:
$\begin{array}{l}
7 \equiv 1\left( {\bmod 3} \right)\\
\Rightarrow {7^n} \equiv 1\left( {\bmod 3} \right)\\
\Rightarrow {7^n} - 1 \equiv 0\left( {\bmod 3} \right)\left( 2 \right)
\end{array}$
Từ $\left( 1 \right),\left( 2 \right) \Rightarrow {11^n} - {2^n} + {7^n} - 1 \equiv 0\left( {\bmod 3} \right)$
$\begin{array}{l}
\Rightarrow {11^n} + {7^n} - {2^n} - 1 \equiv 0\left( {\bmod 3} \right)\\
\Rightarrow \left( {{{11}^n} + {7^n} - {2^n} - 1} \right) \vdots 3\left( * \right)
\end{array}$
+) Mặt khác:
$\begin{array}{l}
11 \equiv 1\left( {\bmod 5} \right)\\
\Rightarrow {11^n} \equiv 1\left( {\bmod 5} \right)\\
\Rightarrow {11^n} - 1 \equiv 0\left( {\bmod 5} \right)\left( 3 \right)
\end{array}$
Và:
$\begin{array}{l}
7 \equiv 2\left( {\bmod 5} \right)\\
\Rightarrow {7^n} \equiv {2^n}\left( {\bmod 5} \right)\\
\Rightarrow {7^n} - {2^n} \equiv 0\left( {\bmod 5} \right)\left( 4 \right)
\end{array}$
Từ $\left( 3 \right),\left( 4 \right) \Rightarrow {11^n} - 1 + {7^n} - {2^n} \equiv 0\left( {\bmod 5} \right)$
$\begin{array}{l}
\Rightarrow {11^n} + {7^n} - {2^n} - 1 \equiv 0\left( {\bmod 5} \right)\\
\Rightarrow \left( {{{11}^n} + {7^n} - {2^n} - 1} \right) \vdots 5\left( {**} \right)
\end{array}$
Như vậy:
Từ $\left( * \right),\left( {**} \right) \Rightarrow \left( {{{11}^n} + {7^n} - {2^n} - 1} \right) \vdots 15\left( {do:\left( {3,5} \right) = 1} \right)$
