گام به گام تمرین صفحه 28 درس 1 حسابان (1) (جبر و معادله)

1)

(الف \(f\left( x \right) = x\left| x \right|\)

\(\left\{ \begin{array}{l}{x^2}\quad ,x \ge 0\\ - {x^2}\;,x < 0\end{array} \right.\)

(ب \(g\left( x \right) = \left| {{x^2} - 1} \right|\)

\(\left\{ \begin{array}{l}{x^2} - 1\quad ,\left| x \right| \ge 1\\ - {x^2} + 1\;,\left| x \right| < 1\end{array} \right.\)

(پ \(h\left( x \right) = \left| {x - 1} \right| + \left| {x + 1} \right|\)

\(\left\{ \begin{array}{l} - 2x\quad ,x <  - 1\\2\begin{array}{*{20}{c}}{}&{}\end{array}\;, - 1 \le x < 1\\2x\quad \;\;\,,1 \le x\end{array} \right.\)

2)

\(\begin{array}{l}\left| {x - \left( { - 1} \right)} \right| + \left| {x - 3} \right| = 6 \Rightarrow \left| {x + 1} \right| + \left| {x - 3} \right| = 6\\\left| {x + 1} \right| + \left| {x - 3} \right| = \left\{ \begin{array}{l} - 2x + 2\quad ,x <  - 1\\4\begin{array}{*{20}{c}}{}&{}&{}\end{array}, - 1 \le x < 3\\2x - 2\quad \;\;\,,3 \le x\end{array} \right.\\\forall \left( {x <  - 1} \right): - 2x + 2 = 6 \Rightarrow {x_1} =  - 2\\\forall \left( {3 \le x} \right):\;\;\;2x - 2 = 6 \Rightarrow {x_2} = 4\end{array}\)

3)

الف) فاصله بین x و 3 برابر 7 است.

\(\begin{array}{l}\left| {x - 3} \right| = 7 \Rightarrow x - 3 =  \pm 7\\ \Rightarrow x = 3 \pm 7\\ \Rightarrow \left\{ \begin{array}{l}{x_1} = 3 + 7 = 10\\{x_2} = 3 - 7 =  - 4\end{array} \right.\end{array}\)

ب) دو برابر فاصله بین x و 6 برابر 4 است.

\(\begin{array}{l}2\left| {x - 6} \right| = 4 \Rightarrow \left| {x - 6} \right| = 2 \Rightarrow x - 6 =  \pm 2\\ \Rightarrow x = 6 \pm 2\\ \Rightarrow \left\{ \begin{array}{l}{x_1} = 6 + 2 = 8\\{x_2} = 6 - 2 = 4\end{array} \right.\end{array}\)

پ) فاصله بین x و 3- بزرگتر از 2 است.

\(\begin{array}{l}\left| {x - \left( { - 3} \right)} \right| > 2 \Rightarrow \left| {x + 3} \right| > 2\\ \Rightarrow \left\{ \begin{array}{l}x + 3 > 2 \Rightarrow x >  - 1\\x + 3 <  - 2 \Rightarrow x <  - 5\end{array} \right.\end{array}\)

4)

(الف \(\frac{{2 - x}}{{\left| {x - 3} \right|}} = 1\)

\(\begin{array}{l} \Rightarrow 2 - x = \left| {x - 3} \right|\;\mathop  \Rightarrow \limits^{{{\left( {} \right)}^2}} \;\\{\left( {2 - x} \right)^2} = {\left| {x - 3} \right|^2} = {\left( {x - 3} \right)^2}\\ \Rightarrow 4 - 4x + {x^2} = {x^2} - 6x + 9\\ \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2}\end{array}\)

(ب \(\sqrt {{x^2} - 2x + 1}  = 2x + 1\)

\(\begin{array}{l}\;\mathop  \Rightarrow \limits^{{{\left( {} \right)}^2}} \;{\left( {\sqrt {{x^2} - 2x + 1} } \right)^2} = {\left( {2x + 1} \right)^2}\\ \Rightarrow {x^2} - 2x + 1 = 4{x^2} + 4x + 1\\ \Rightarrow 3{x^2} + 6x = 0 \Rightarrow 3x\left( {x + 2} \right) = 0\\ \Rightarrow \left\{ \begin{array}{l}3x = 0 \Rightarrow {x_1} = 0\\x + 2 = 0 \Rightarrow {x_2} =  - 2\;\; \otimes \\ \Rightarrow 3 = \sqrt {{{\left( { - 2} \right)}^2} - 2\left( { - 2} \right) + 1}  \ne 2\left( { - 2} \right) + 1 =  - 3\end{array} \right.\end{array}\)

5)

(الف \(y = x - \frac{x}{{\left| x \right|}}\)

\(\left\{ \begin{array}{l}x - 1\quad ,x > 0\\x + 1\quad ,x < 0\end{array} \right.\)

روش جبری :

\(\left\{ \begin{array}{l}\forall \left( {x > 0} \right):\quad x - 1 = 3 \Rightarrow x = 4\\\forall \left( {x < 0} \right):\quad x + 1 = 3 \Rightarrow x = 2\; \otimes \end{array} \right. \Rightarrow {x_ \circ } = 4\)

روش هندسی :  \({x_ \circ } = 4\)

(ب \(y = \left| {{x^2} - 6x} \right|\)

\(\left\{ \begin{array}{l}{x^2} - 6x\quad \;\;,x < 0\\ - {x^2} + 6x\quad ,0 \le x < 6\\{x^2} - 6x\quad \;\;,6 \le x\end{array} \right.\)

روش جبری :

\(\begin{array}{l}\forall \left( {x < 0} \right) \cup \left( {6 \le x} \right):\quad {x^2} - 6x = 3 \Rightarrow \left\{ \begin{array}{l}{x_1} = 3 - 2\sqrt 3 \\{x_2} = 3 + 2\sqrt 3 \end{array} \right.\\\forall \left( {0 \le x < 6} \right):\quad  - {x^2} + 6x = 3 \Rightarrow \left\{ \begin{array}{l}{x_3} = 3 - \sqrt 6 \\{x_4} = 3 + \sqrt 6 \end{array} \right.\end{array}\)

روش هندسی :

\(\left\{ \begin{array}{l}{x_1} =  - 0/5\quad ,\quad {x_2} = 0/5\\{x_3} = 5/5\quad ,\quad {x_4} = 6/5\end{array} \right.\)

6)

: روش جبری

\(\left| {\left| x \right| - 2} \right| = 1 \Rightarrow \left\{ \begin{array}{l}\left| x \right| - 2 = 1 \Rightarrow \left| x \right| = 3 \Rightarrow \left\{ \begin{array}{l}{x_1} = 3\\{x_2} =  - 3\end{array} \right.\\\left| x \right| - 2 =  - 1 \Rightarrow \left| x \right| = 1 \Rightarrow \left\{ \begin{array}{l}{x_3} = 1\\{x_4} =  - 1\end{array} \right.\end{array} \right.\)

7)

: روش جبری

\(\begin{array}{l}\left| {{x^2} - 2x} \right| = 2\\\forall \left( {x < 0} \right) \cup \left( {2 \le x} \right):\quad {x^2} - 2x = 2 \Rightarrow \left\{ \begin{array}{l}{x_1} = 1 - \sqrt 3 \\{x_2} = 1 + \sqrt 3 \end{array} \right.\\\forall \left( {0 \le x < 2} \right):\quad  - {x^2} + 2x = 2 \Rightarrow  - {x^2} + 2x - 2 = 0\\\begin{array}{*{20}{c}}{}&{}&{}&{}&{}\end{array}\;\Delta  = {\left( 2 \right)^2} - 4\left( { - 1} \right)\left( { - 2} \right) =  - 4 < 0\end{array}\)