B= x+2/3-x. Tìm x để B lớn hơn 0, B<0 C=(x-1)(x+2)(3-x). Tìm x để C<0, C lớn hơn hoặc bằng 0 giải theo cách lớp 7 nhaa

Đáp án:

\(\begin{array}{l}
b)C < 0\\
 \to \left[ \begin{array}{l}
x > 3\\
 - 1 > x >  - 2
\end{array} \right.\\
C \ge 0\\
 \to \left[ \begin{array}{l}
3 \ge x \ge 1\\
x \le  - 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne 3\\
B > 0\\
 \to \dfrac{{x + 2}}{{3 - x}} > 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
3 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
3 - x < 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3 > x >  - 2\\
\left\{ \begin{array}{l}
x <  - 2\\
3 < x
\end{array} \right.\left( {KTM} \right)
\end{array} \right.\\
 \to 3 > x >  - 2\\
B < 0\\
 \to \dfrac{{x + 2}}{{3 - x}} < 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
3 - x < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
3 - x > 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x >  - 2\\
3 < x
\end{array} \right.\\
\left\{ \begin{array}{l}
x <  - 2\\
3 > x
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x > 3\\
x <  - 2
\end{array} \right.\\
b)C < 0\\
TH1:x - 1 > 0 \to x > 1\\
 \to \left( {x + 2} \right)\left( {3 - x} \right) < 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
3 - x < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
3 - x > 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x >  - 2\\
3 < x
\end{array} \right.\\
\left\{ \begin{array}{l}
x <  - 2\\
3 > x
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x > 3\\
x <  - 2
\end{array} \right.\\
 \to x > 3\\
TH2:x - 1 < 0 \to x < 1\\
 \to \left( {x + 2} \right)\left( {3 - x} \right) > 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
3 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
3 - x < 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3 > x >  - 2\\
\left\{ \begin{array}{l}
x <  - 2\\
3 < x
\end{array} \right.\left( {KTM} \right)
\end{array} \right.\\
 \to 3 > x >  - 2\\
 \to  - 1 > x >  - 2\\
KL:\left[ \begin{array}{l}
x > 3\\
 - 1 > x >  - 2
\end{array} \right.\\
C \ge 0\\
TH1:x - 1 \ge 0 \to x \ge 1\\
 \to \left( {x + 2} \right)\left( {3 - x} \right) \ge 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 \ge 0\\
3 - x \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 \le 0\\
3 - x \le 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
3 \ge x \ge  - 2\\
\left\{ \begin{array}{l}
x \le  - 2\\
3 \le x
\end{array} \right.\left( {KTM} \right)
\end{array} \right.\\
 \to 3 \ge x \ge  - 2\\
 \to 3 \ge x \ge 1\\
TH2:x - 1 \le 0 \to x \le 1\\
 \to \left( {x + 2} \right)\left( {3 - x} \right) \le 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 \ge 0\\
3 - x \le 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 \le 0\\
3 - x \ge 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge  - 2\\
3 \le x
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le  - 2\\
3 \ge x
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x \ge 3\\
x \le  - 2
\end{array} \right.\\
 \to x \le  - 2\\
KL:\left[ \begin{array}{l}
3 \ge x \ge 1\\
x \le  - 2
\end{array} \right.
\end{array}\)