Giải các bpt sau:
a) (x^2–1)(4-2x)(5x-3)≤0
b) (x-2x+1)(3-x) /4x-16≤0
c) 1/x+2/x+4<3/x+3
Giải các bpt sau: a) (x^2–1)(4-2x)(5x-3)≤0 b) (x-2x+1)(3-x) /4x-16≤0 c) 1/x+2/x+4<3/x+3
By Maria
By Maria
Giải các bpt sau:
a) (x^2–1)(4-2x)(5x-3)≤0
b) (x-2x+1)(3-x) /4x-16≤0
c) 1/x+2/x+4<3/x+3
Đáp án:
a) \(x \in \left( { – \infty ; – 1} \right] \cup \left[ {\dfrac{3}{5};1} \right] \cup \left[ {2; + \infty } \right)\)
Giải thích các bước giải:
\(a)({x^2} – 1)(4 – 2x)(5x – 3) \le 0\)
BXD:
x -∞ -1 3/5 1 2 +∞
f(x) – 0 + 0 – 0 + 0 –
\(KL:x \in \left( { – \infty ; – 1} \right] \cup \left[ {\dfrac{3}{5};1} \right] \cup \left[ {2; + \infty } \right)\)
\(\begin{array}{l}
b)DK:x \ne 4\\
\dfrac{{\left( {{x^2} – 2x + 1} \right)\left( {3 – x} \right)}}{{4x – 16}} \le 0\\
\to \dfrac{{{{\left( {x – 1} \right)}^2}\left( {3 – x} \right)}}{{4\left( {x – 4} \right)}} \le 0
\end{array}\)
BXD:
x -∞ 1(kép) 3 4 +∞
f(x) – 0 – 0 + // –
\(KL:x \in \left( { – \infty ;3} \right] \cup \left( {4; + \infty } \right)\)
\(\begin{array}{l}
c)DK:x \ne \left\{ { – 4; – 3;0} \right\}\\
\dfrac{1}{x} + \dfrac{2}{{x + 4}} < \dfrac{3}{{x + 3}}\\
\to \dfrac{{{x^2} + 7x + 12 + 2{x^2} + 6x – 3{x^2} – 12x}}{{x\left( {x + 3} \right)\left( {x + 4} \right)}} < 0\\
\to \dfrac{{x + 12}}{{x\left( {x + 3} \right)\left( {x + 4} \right)}} < 0
\end{array}\)
BXD:
x -∞ -12 -4 -3 0 +∞
f(x) + 0 – // + // – // +
\(KL:x \in \left( { – 12; – 4} \right) \cup \left( { – 3;0} \right)\)