Đáp án:
b) $B = \dfrac{{21}}{2}$ khi $x=4$
c) $x \in \left\{ {3;9} \right\}$
d)$x \in \left\{ {\dfrac{1}{4}; - 2} \right\}$
e) $x \in \left( { - \infty ;2} \right)\backslash \left\{ { - 5;1} \right\}$
f)$MaxM = \dfrac{8}{3} \Leftrightarrow x = \dfrac{{ - 1}}{2}$
g)$MinB = 5 + 2\sqrt 7 \Leftrightarrow x = 2 + \sqrt 7 $
Giải thích các bước giải:
ĐKXĐ: $x \ne \left\{ { - 5;1;2} \right\}$
a) Ta có:
$\begin{array}{l}
B = \left( {\dfrac{{9 - 3x}}{{{x^2} + 4x - 5}} - \dfrac{{x + 5}}{{1 - x}} - \dfrac{{x + 1}}{{x + 5}}} \right):\dfrac{{7x - 14}}{{{x^3} - 1}}\\
= \left( {\dfrac{{9 - 3x}}{{\left( {x - 1} \right)\left( {x + 5} \right)}} + \dfrac{{x + 5}}{{x - 1}} - \dfrac{{x + 1}}{{x + 5}}} \right):\dfrac{{7\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{9 - 3x + {{\left( {x + 5} \right)}^2} - \left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 5} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{7\left( {x - 2} \right)}}\\
= \dfrac{{7x + 35}}{{\left( {x - 1} \right)\left( {x + 5} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{7\left( {x - 2} \right)}}\\
= \dfrac{{{x^2} + x + 1}}{{x - 2}}
\end{array}$
b) Ta có:
$\begin{array}{l}
{\left( {x + 5} \right)^2} - 9x - 45 = 0\\
\Leftrightarrow {x^2} + x - 20 = 0\\
\Leftrightarrow \left( {x + 5} \right)\left( {x - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 5\left( l \right)\\
x = 4\left( c \right)
\end{array} \right.\\
\Leftrightarrow x = 4
\end{array}$
Vậy với $x=4$ thì $B = \dfrac{{{x^2} + x + 1}}{{x - 2}} = \dfrac{{{4^2} + 4 + 1}}{{4 - 2}} = \dfrac{{21}}{2}$
Vậy $B = \dfrac{{21}}{2}$ khi $x=4$
c) Ta có:
$\begin{array}{l}
B \in Z\\
\Leftrightarrow \dfrac{{{x^2} + x + 1}}{{x - 2}} \in Z\\
\Leftrightarrow \dfrac{{\left( {x - 2} \right)\left( {x + 3} \right) + 7}}{{x - 2}} \in Z\\
\Leftrightarrow x + 3 + \dfrac{7}{{x - 2}} \in Z\\
\Leftrightarrow \dfrac{7}{{x - 2}} \in Z\\
\Leftrightarrow \left( {x - 2} \right) \in U\left( 7 \right) = \left\{ { - 7; - 1;1;7} \right\}\\
\Leftrightarrow x \in \left\{ { - 5;1;3;9} \right\}\\
\Leftrightarrow x \in \left\{ {3;9} \right\}\left( {Do:x\not \in \left\{ { - 1;5} \right\}} \right)
\end{array}$
Vậy $x \in \left\{ {3;9} \right\}$ thỏa mãn
d) Ta có;
$\begin{array}{l}
B = \dfrac{{ - 3}}{4}\\
\Leftrightarrow \dfrac{{{x^2} + x + 1}}{{x - 2}} = \dfrac{{ - 3}}{4}\\
\Leftrightarrow 4\left( {{x^2} + x + 1} \right) + 3\left( {x - 2} \right) = 0\\
\Leftrightarrow 4{x^2} + 7x - 2 = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {4x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 2\\
x = \dfrac{1}{4}
\end{array} \right.\left( c \right)
\end{array}$
Vậy $x \in \left\{ {\dfrac{1}{4}; - 2} \right\}$
e) Ta có:
$\begin{array}{l}
B < 0\\
\Leftrightarrow \dfrac{{{x^2} + x + 1}}{{x - 2}} < 0\\
\Leftrightarrow x - 2 < 0\left( {do:{x^2} + x + 1 = {{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4} > 0,\forall x} \right)\\
\Leftrightarrow x < 2
\end{array}$
Kết hợp với ĐKXĐ ta có: $x \in \left( { - \infty ;2} \right)\backslash \left\{ { - 5;1} \right\}$ thỏa mãn đề.
Vậy $x \in \left( { - \infty ;2} \right)\backslash \left\{ { - 5;1} \right\}$
f) Ta có
$\begin{array}{l}
M = \dfrac{2}{{x - 2}}:B\\
= \dfrac{2}{{x - 2}}:\dfrac{{{x^2} + x + 1}}{{x - 2}}\\
= \dfrac{2}{{{x^2} + x + 1}}
\end{array}$
Mà lại có:
$\begin{array}{l}
{x^2} + x + 1 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4},\forall x\\
\Rightarrow \dfrac{2}{{{x^2} + x + 1}} \le \dfrac{2}{{\dfrac{3}{4}}} = \dfrac{8}{3}\\
\Rightarrow M \le \dfrac{8}{3}
\end{array}$
Dấu bằng xảy ra$ \Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = 0 \Leftrightarrow x = \dfrac{{ - 1}}{2}$
Vậy $MaxM = \dfrac{8}{3} \Leftrightarrow x = \dfrac{{ - 1}}{2}$
g) Ta có:
Với $x > 2$ ta có $x-2>0$
Khi đó:
$\begin{array}{l}
B = \dfrac{{{x^2} + x + 1}}{{x - 2}}\\
= x + 3 + \dfrac{7}{{x - 2}}\\
= \left( {x - 2 + \dfrac{7}{{x - 2}}} \right) + 5\\
\ge 2\sqrt {\left( {x - 2} \right).\dfrac{7}{{x - 2}}} + 5\left( {BDT:Cauchy} \right)\\
= 5 + 2\sqrt 7
\end{array}$
Dấu bằng xảy ra $ \Leftrightarrow x - 2 = \dfrac{7}{{x - 2}} \Leftrightarrow x - 2 = \sqrt 7 \Leftrightarrow x = 2 + \sqrt 7 $
Vậy $MinB = 5 + 2\sqrt 7 \Leftrightarrow x = 2 + \sqrt 7 $
