Giải bpt: a, 2x^2 + 3x >= 3.[căn(2x^2 + 3x+9)] +9 b, căn(-x^2 + 4x – 3) < 2x -5 c, gttđ(x-2) < 2x -3 22/09/2021 Bởi Margaret Giải bpt: a, 2x^2 + 3x >= 3.[căn(2x^2 + 3x+9)] +9 b, căn(-x^2 + 4x – 3) < 2x -5 c, gttđ(x-2) < 2x -3
Đáp án: c. \(x \in \left( {\dfrac{5}{3}; + \infty } \right)\) Giải thích các bước giải: \(\begin{array}{l}a.2{x^2} + 3x \ge 3\sqrt {2{x^2} + 3x + 9} + 9\\ \to 2{x^2} + 3x – 9 \ge 3\sqrt {2{x^2} + 3x + 9} \left( 1 \right)\\Đặt:\sqrt {2{x^2} + 3x + 9} = t\left( {t \ge 0} \right)\\ \to 2{x^2} + 3x + 9 = {t^2}\\ \to 2{x^2} + 3x = {t^2} – 9\\\left( 1 \right) \to {t^2} – 9 – 9 = 3t\\ \to {t^2} – 3t – 18 = 0\\ \to \left( {t – 6} \right)\left( {t + 3} \right) = 0\\ \to \left[ \begin{array}{l}t = 6\\t = – 3\left( l \right)\end{array} \right.\\ \to \sqrt {2{x^2} + 3x + 9} = 6\\ \to 2{x^2} + 3x + 9 = 36\\ \to \left[ \begin{array}{l}x = 3\\x = – \dfrac{9}{2}\end{array} \right.\\b.\sqrt { – {x^2} + 4x – 3} < 2x – 5\\ \to \left\{ \begin{array}{l} – {x^2} + 4x – 3 \ge 0\\2x – 5 \ge 0\\ – {x^2} + 4x – 3 < 4{x^2} – 20x + 25\end{array} \right.\\ \to \left\{ \begin{array}{l}x \in \left[ {1;3} \right]\\x \ge \dfrac{5}{2}\\5{x^2} – 24x + 28 > 0\end{array} \right.\\ \to \left\{ \begin{array}{l}x \in \left[ {\dfrac{5}{2};3} \right]\\x \in \left( { – \infty ;2} \right) \cup \left( {\dfrac{{14}}{5}; + \infty } \right)\end{array} \right.\\ \to x \in \left( {\dfrac{{14}}{5};3} \right]\\c.\left| {x – 2} \right| < 2x – 3\\ \to \left\{ \begin{array}{l}{x^2} – 4x + 4 < 4{x^2} – 12x + 9\\2x – 3 \ge 0\end{array} \right.\\ \to \left\{ \begin{array}{l}3{x^2} – 8x + 5 > 0\\x \ge \dfrac{3}{2}\end{array} \right.\\ \to \left\{ \begin{array}{l}x \in \left( { – \infty ;1} \right) \cup \left( {\dfrac{5}{3}; + \infty } \right)\\x \ge \dfrac{3}{2}\end{array} \right.\\ \to x \in \left( {\dfrac{5}{3}; + \infty } \right)\end{array}\) Bình luận
Đáp án:
c. \(x \in \left( {\dfrac{5}{3}; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a.2{x^2} + 3x \ge 3\sqrt {2{x^2} + 3x + 9} + 9\\
\to 2{x^2} + 3x – 9 \ge 3\sqrt {2{x^2} + 3x + 9} \left( 1 \right)\\
Đặt:\sqrt {2{x^2} + 3x + 9} = t\left( {t \ge 0} \right)\\
\to 2{x^2} + 3x + 9 = {t^2}\\
\to 2{x^2} + 3x = {t^2} – 9\\
\left( 1 \right) \to {t^2} – 9 – 9 = 3t\\
\to {t^2} – 3t – 18 = 0\\
\to \left( {t – 6} \right)\left( {t + 3} \right) = 0\\
\to \left[ \begin{array}{l}
t = 6\\
t = – 3\left( l \right)
\end{array} \right.\\
\to \sqrt {2{x^2} + 3x + 9} = 6\\
\to 2{x^2} + 3x + 9 = 36\\
\to \left[ \begin{array}{l}
x = 3\\
x = – \dfrac{9}{2}
\end{array} \right.\\
b.\sqrt { – {x^2} + 4x – 3} < 2x – 5\\
\to \left\{ \begin{array}{l}
– {x^2} + 4x – 3 \ge 0\\
2x – 5 \ge 0\\
– {x^2} + 4x – 3 < 4{x^2} – 20x + 25
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left[ {1;3} \right]\\
x \ge \dfrac{5}{2}\\
5{x^2} – 24x + 28 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left[ {\dfrac{5}{2};3} \right]\\
x \in \left( { – \infty ;2} \right) \cup \left( {\dfrac{{14}}{5}; + \infty } \right)
\end{array} \right.\\
\to x \in \left( {\dfrac{{14}}{5};3} \right]\\
c.\left| {x – 2} \right| < 2x – 3\\
\to \left\{ \begin{array}{l}
{x^2} – 4x + 4 < 4{x^2} – 12x + 9\\
2x – 3 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3{x^2} – 8x + 5 > 0\\
x \ge \dfrac{3}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { – \infty ;1} \right) \cup \left( {\dfrac{5}{3}; + \infty } \right)\\
x \ge \dfrac{3}{2}
\end{array} \right.\\
\to x \in \left( {\dfrac{5}{3}; + \infty } \right)
\end{array}\)