Toán Biểu thức f(x)=1/x+1 + 2/2-x dương khi và chỉ khi x thuộc ? 06/10/2021 By Piper Biểu thức f(x)=1/x+1 + 2/2-x dương khi và chỉ khi x thuộc ?
Đáp án: $x \in \left( { – 1;2} \right) \cup \left( { – \infty ; – 4} \right)$ Giải thích các bước giải: ĐKXĐ $x\ne -1;x\ne 2$ Ta có: $f\left( x \right) = \dfrac{1}{{x + 1}} + \dfrac{2}{{2 – x}}$ Để $f\left( x \right) > 0$ $\begin{array}{l} \Leftrightarrow \dfrac{1}{{x + 1}} + \dfrac{2}{{2 – x}} > 0\\ \Leftrightarrow \dfrac{{2 – x + 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {2 – x} \right)}} > 0\\ \Leftrightarrow \dfrac{{x + 4}}{{\left( {x + 1} \right)\left( {2 – x} \right)}} > 0\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + 4 > 0\\\left( {x + 1} \right)\left( {2 – x} \right) > 0\end{array} \right.\\\left\{ \begin{array}{l}x + 4 < 0\\\left( {x + 1} \right)\left( {2 – x} \right) < 0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > – 4\\\left( {x + 1} \right)\left( {x – 2} \right) < 0\end{array} \right.\\\left\{ \begin{array}{l}x < – 4\\\left( {x + 1} \right)\left( {x – 2} \right) > 0\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x > – 4\\ – 1 < x < 2\end{array} \right.\\\left\{ \begin{array}{l}x < – 4\\\left[ \begin{array}{l}x > 2\\x < – 1\end{array} \right.\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} – 1 < x < 2\\x < – 4\end{array} \right.\\ \Leftrightarrow x \in \left( { – 1;2} \right) \cup \left( { – \infty ; – 4} \right)\end{array}$ Vậy $x \in \left( { – 1;2} \right) \cup \left( { – \infty ; – 4} \right)$ để $f\left( x \right) > 0$ Trả lời
Đáp án:
$x \in \left( { – 1;2} \right) \cup \left( { – \infty ; – 4} \right)$
Giải thích các bước giải:
ĐKXĐ $x\ne -1;x\ne 2$
Ta có:
$f\left( x \right) = \dfrac{1}{{x + 1}} + \dfrac{2}{{2 – x}}$
Để $f\left( x \right) > 0$
$\begin{array}{l}
\Leftrightarrow \dfrac{1}{{x + 1}} + \dfrac{2}{{2 – x}} > 0\\
\Leftrightarrow \dfrac{{2 – x + 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {2 – x} \right)}} > 0\\
\Leftrightarrow \dfrac{{x + 4}}{{\left( {x + 1} \right)\left( {2 – x} \right)}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 4 > 0\\
\left( {x + 1} \right)\left( {2 – x} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 4 < 0\\
\left( {x + 1} \right)\left( {2 – x} \right) < 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > – 4\\
\left( {x + 1} \right)\left( {x – 2} \right) < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < – 4\\
\left( {x + 1} \right)\left( {x – 2} \right) > 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > – 4\\
– 1 < x < 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x < – 4\\
\left[ \begin{array}{l}
x > 2\\
x < – 1
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
– 1 < x < 2\\
x < – 4
\end{array} \right.\\
\Leftrightarrow x \in \left( { – 1;2} \right) \cup \left( { – \infty ; – 4} \right)
\end{array}$
Vậy $x \in \left( { – 1;2} \right) \cup \left( { – \infty ; – 4} \right)$ để $f\left( x \right) > 0$
.