Xét dấu các biểu thức sau:
a) f(x)= (x-2).(x+1)/3x -1
b) f(x)= 1-4x /2x-3
c) f(x)= 3x (6 -2x)/5x-4
d) f(x)= -4/3x+1 – 3/2-x
e) f(x)= 2x-5/(3x+1) (2-4x)
Xét dấu các biểu thức sau:
a) f(x)= (x-2).(x+1)/3x -1
b) f(x)= 1-4x /2x-3
c) f(x)= 3x (6 -2x)/5x-4
d) f(x)= -4/3x+1 – 3/2-x
e) f(x)= 2x-5/(3x+1) (2-4x)
Đáp án:
b) \(\begin{array}{l}
f\left( x \right) > 0 \Leftrightarrow x \in \left( {\dfrac{1}{4};\dfrac{3}{2}} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { – \infty ;\dfrac{1}{4}} \right) \cup \left( {\dfrac{3}{2}; + \infty } \right)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \dfrac{1}{3}\\
Xét:f\left( x \right) = 0\\
\to \dfrac{{\left( {x – 2} \right)\left( {x + 1} \right)}}{{3x – 1}} = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = – 1
\end{array} \right.
\end{array}\)
BXD:
x -∞ -1 1/3 2 +∞
f(x) – 0 + // – 0 +
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( { – 1;\dfrac{1}{3}} \right) \cup \left( {2; + \infty } \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { – \infty ; – 1} \right) \cup \left( {\dfrac{1}{3};2} \right)
\end{array}\)
\(b)DK:x \ne \dfrac{3}{2}\)
BXD:
x -∞ 1/4 3/2 +∞
f(x) – 0 + // –
\(\begin{array}{l}
KL:f\left( x \right) > 0 \Leftrightarrow x \in \left( {\dfrac{1}{4};\dfrac{3}{2}} \right)\\
f\left( x \right) < 0 \Leftrightarrow x \in \left( { – \infty ;\dfrac{1}{4}} \right) \cup \left( {\dfrac{3}{2}; + \infty } \right)
\end{array}\)
\(c)DK:x \ne \dfrac{4}{5}\)
BXD:
x -∞ 0 4/5 3 +∞
f(x) + 0 – // + 0 –
\(\begin{array}{l}
d)DK:x \ne \left\{ { – \dfrac{1}{3};2} \right\}\\
f\left( x \right) = – \dfrac{4}{{3x + 1}} – \dfrac{3}{{2 – x}}\\
= \dfrac{{ – 8 + 4x – 9x – 3}}{{\left( {2 – x} \right)\left( {3x + 1} \right)}}\\
= \dfrac{{ – 11 – 5x}}{{\left( {2 – x} \right)\left( {3x + 1} \right)}}
\end{array}\)
BXD:
x -∞ -11/5 -1/3 2 +∞
f(x) – 0 + // – // +
\(e)DK:x \ne \left\{ { – \dfrac{1}{2}; – \dfrac{1}{3}} \right\}\)
BXD:
x -∞ -1/2 -1/3 5/2 +∞
f(x) + // – // + 0 –