Giải các bất phương trình: a, x- 2x+1/2- x+2/3 > 1 b, (2x+3)(2x-1) < (2x-5)^2 05/10/2021 Bởi Ximena Giải các bất phương trình: a, x- 2x+1/2- x+2/3 > 1 b, (2x+3)(2x-1) < (2x-5)^2
Đáp án: $\begin{array}{l}a)x – \dfrac{{2x + 1}}{2} – \dfrac{{x + 2}}{3} > 1\\ \Rightarrow \dfrac{{6x – 3\left( {2x + 1} \right) – 2\left( {x + 2} \right)}}{6} > 1\\ \Rightarrow \dfrac{{6x – 6x – 3 – 2x – 4 – 6}}{6} > 0\\ \Rightarrow – 2x – 13 > 0\\ \Rightarrow x < \dfrac{{ – 13}}{2}\\Vậy\,x < \dfrac{{ – 13}}{2}\\b)\left( {2x + 3} \right)\left( {2x – 1} \right) < {\left( {2x – 5} \right)^2}\\ \Rightarrow 4{x^2} – 2x + 6x – 3 < 4{x^2} – 20x + 25\\ \Rightarrow 4{x^2} + 4x – 3 – 4{x^2} + 20x – 25 < 0\\ \Rightarrow 24x – 28 < 0\\ \Rightarrow 24x < 28\\ \Rightarrow x < \dfrac{7}{6}\\Vậy\,x < \dfrac{7}{6}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)x – \dfrac{{2x + 1}}{2} – \dfrac{{x + 2}}{3} > 1\\
\Rightarrow \dfrac{{6x – 3\left( {2x + 1} \right) – 2\left( {x + 2} \right)}}{6} > 1\\
\Rightarrow \dfrac{{6x – 6x – 3 – 2x – 4 – 6}}{6} > 0\\
\Rightarrow – 2x – 13 > 0\\
\Rightarrow x < \dfrac{{ – 13}}{2}\\
Vậy\,x < \dfrac{{ – 13}}{2}\\
b)\left( {2x + 3} \right)\left( {2x – 1} \right) < {\left( {2x – 5} \right)^2}\\
\Rightarrow 4{x^2} – 2x + 6x – 3 < 4{x^2} – 20x + 25\\
\Rightarrow 4{x^2} + 4x – 3 – 4{x^2} + 20x – 25 < 0\\
\Rightarrow 24x – 28 < 0\\
\Rightarrow 24x < 28\\
\Rightarrow x < \dfrac{7}{6}\\
Vậy\,x < \dfrac{7}{6}
\end{array}$