Tìm \(x\), biết :a) \(\left| {3x - 5} \right| + \left| {2x + 3} \right| = 7\) b) \(\left| x \right| + \left|

Câu hỏi số 617739:

Thông hiểu

Tìm \(x\), biết :

a) \(\left| {3x - 5} \right| + \left| {2x + 3} \right| = 7\)

b) \(\left| x \right| + \left| {x + 2} \right| = 3\)

c) \(2\left| {x - 3} \right| + \left| {2x - 5} \right| = 11\)

d) \(\left| {2x + 3} \right| - 2\left| {4 - x} \right| = 5\)

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Xem lời giải

Câu hỏi:617739

Phương pháp giải

+ Định lí dấu nhị thức bậc nhất: \(ax + b\left( {a \ne 0} \right)\)

Giả sử \({x_0}\) là nghiệm của \(ax + b\). Khi đó:

Nhị thức cùng dấu với \(a\) nếu \(x > {x_0}\)
Nhị thức trái dấu với \(a\) nếu \(x < {x_0}\)

+ Nếu \(x - a \ge 0 \Rightarrow \left| {x - a} \right| = x - a\)

\(x - a \le 0 \Rightarrow \left| {x - a} \right| = a - x\)

Giải chi tiết

a) \(\left| {3x - 5} \right| + \left| {2x + 3} \right| = 7\left( 1 \right)\)

Bảng xét dấu:

+ Nếu \(x < - \dfrac{3}{2}\) thì \(\left\{ \begin{array}{l}3x - 5 < 0\\2x + 3 < 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {3x - 5} \right| = - \left( {3x - 5} \right) = - 3x + 5\\\left| {2x + 3} \right| = - \left( {2x + 3} \right) = - 2x - 3\end{array} \right.\)

\(\left( 1 \right) \Rightarrow - 3x + 5 - 2x - 3 = 7\)

\(\begin{array}{l} - 5x = 5\\\,\,\,\,\,x = - 1\left( {{\rm{ktm}}\,x < - \dfrac{3}{2}} \right)\end{array}\)

+ Nếu \( - \dfrac{3}{2} \le x \le \dfrac{5}{3}\) thì \(\left\{ \begin{array}{l}3x - 5 \le 0\\2x + 3 \ge 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {3x - 5} \right| = - \left( {3x - 5} \right) = - 3x + 5\\\left| {2x + 3} \right| = 2x + 3\end{array} \right.\)

\(\left( 1 \right) \Rightarrow - 3x + 5 + 2x + 3 = 7\)

\(\begin{array}{l} - x = - 1\\\,\,\,x = 1\left( {tm} \right)\end{array}\)

+ Nếu \(x > \dfrac{5}{3}\) thì \(\left\{ \begin{array}{l}3x - 5 > 0\\2x + 3 > 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {3x - 5} \right| = 3x - 5\\\left| {2x + 3} \right| = 2x + 3\end{array} \right.\)

\(\left( 1 \right) \Rightarrow 3x - 5 + 2x - 3 = 7\)

\(\begin{array}{l}5x = 15\\\,\,x = 3\left( {tm} \right)\end{array}\)

Vậy \(x \in \left\{ {1;3} \right\}\)

b) \(\left| x \right| + \left| {x + 2} \right| = 3\left( 2 \right)\)

Bảng xét dấu

+ Nếu \(x < - 2\) thì \(\left\{ \begin{array}{l}x < 0\\x + 2 < 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| x \right| = - x\\\left| {x + 2} \right| = - \left( {x + 2} \right) = - x - 2\end{array} \right.\)

\(\left( 2 \right) \Rightarrow - x - x - 2 = 3\)

\(\begin{array}{l} - 2x = 5\\\,\,\,\,\,\,x = - \dfrac{5}{2}\left( {tm} \right)\end{array}\)

+ Nếu \( - 2 \le x \le 0\) thì \(\left\{ \begin{array}{l}x \le 0\\x + 2 \ge 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| x \right| = - x\\\left| {x + 2} \right| = x + 2\end{array} \right.\)

\(\left( 2 \right) \Rightarrow - x + x + 2 = 3\)

\(2 = 3\) (vô lí)

+ Nếu \(x > 0\) thì \(\left\{ \begin{array}{l}x > 0\\x + 2 > 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| x \right| = x\\\left| {x + 2} \right| = x + 2\end{array} \right.\)

\(\left( 2 \right) \Rightarrow x + x + 2 = 3\)

\(\begin{array}{l}2x = 1\\\,\,\,x = \dfrac{1}{2}\left( {tm} \right)\end{array}\)

Vậy \(x \in \left\{ { - \dfrac{5}{2};\dfrac{1}{2}} \right\}\)

c) \(2\left| {x - 3} \right| + \left| {2x - 5} \right| = 11\left( 3 \right)\)

Bảng xét dấu

+ Nếu \(x < \dfrac{5}{2}\) thì \(\left\{ \begin{array}{l}x - 3 < 0\\2x - 5 < 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {x - 3} \right| = - \left( {x - 3} \right) = - x + 3\\\left| {2x - 5} \right| = - \left( {2x - 5} \right) = - 2x + 5\end{array} \right.\)

\(\left( 3 \right) \Rightarrow 2\left( { - x + 3} \right) - 2x + 5 = 11\)

\(\begin{array}{l} - 2x + 6 - 2x + 5 = 11\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 4x = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0\left( {tm} \right)\end{array}\)

+ Nếu \(\dfrac{5}{2} \le x \le 3\) thì \(\left\{ \begin{array}{l}x - 3 \le 0\\2x - 5 \ge 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {x - 3} \right| = - \left( {x - 3} \right) = - x + 3\\\left| {2x - 5} \right| = 2x - 5\end{array} \right.\)

\(\left( 3 \right) \Rightarrow 2\left( { - x + 3} \right) + 2x - 5 = 11\)

\( - 2x + 6 + 2x - 5 = 11\)

\(1 = 11\) (vô lí)

+ Nếu \(x > 3\) thì \(\left\{ \begin{array}{l}x - 3 > 0\\2x - 5 > 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {x - 3} \right| = x - 3\\\left| {2x - 5} \right| = 2x - 5\end{array} \right.\)

\(\left( 3 \right) \Rightarrow 2\left( {x - 3} \right) + 2x - 5 = 11\)

\(\begin{array}{l}2x - 6 + 2x - 5 = 11\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4x = 22\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \dfrac{{11}}{2}\left( {tm} \right)\end{array}\)

Vậy \(x \in \left\{ {0;\dfrac{{11}}{2}} \right\}\)

d) \(\left| {2x + 3} \right| - 2\left| {4 - x} \right| = 5\left( 4 \right)\)

Bảng xét dấu:

+ Nếu \(x < - 4\) thì \(\left\{ \begin{array}{l}2x + 3 < 0\\4 - x < 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {2x + 3} \right| = - \left( {2x + 3} \right) = - 2x - 3\\\left| {4 - x} \right| = - \left( {4 - x} \right) = x - 4\end{array} \right.\)

\(\left( 4 \right) \Rightarrow - 2x - 3 - 2\left( {x - 4} \right) = 5\)

\(\begin{array}{l} - 2x - 3 - 2x + 8 = 5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 4x = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0\left( {ktm\,\,\,x < - 4} \right)\end{array}\)

+ Nếu \( - 4 \le x \le - \dfrac{3}{2}\) thì \(\left\{ \begin{array}{l}2x + 3 \le 0\\4 - x \ge 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {2x + 3} \right| = - \left( {2x + 3} \right) = - 2x - 3\\\left| {4 - x} \right| = 4 - x\end{array} \right.\)

\(\left( 4 \right) \Rightarrow - 2x - 3 - 2\left( {4 - x} \right) = 5\)

\( - 2x - 3 - 8 + 2x = 5\)

\( - 11 = 5\) (vô lí)

+ Nếu \(x > - \dfrac{3}{2}\) thì \(\left\{ \begin{array}{l}2x + 3 > 0\\4 - x > 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left| {2x + 3} \right| = 2x + 3\\\left| {4 - x} \right| = 4 - x\end{array} \right.\)

\(\left( 4 \right) \Rightarrow 2x + 3 - 2\left( {4 - x} \right) = 5\)

\(\begin{array}{l}2x + 3 - 8 + 2x = 5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4x = 10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = \dfrac{5}{2}\left( {tm} \right)\end{array}\)

Vậy \(x \in \left\{ {0;\dfrac{5}{2}} \right\}\)

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