گام به گام تمرین صفحه 22 درس 1 حسابان (1) (جبر و معادله)

1) \(\frac{6}{x} = 2 + \frac{{x - 3}}{{x + 1}}\)

\(\begin{array}{l}\frac{6}{x} = 2 + \frac{{x - 3}}{{x + 1}}\;\;\;\mathop  \Rightarrow \limits^{ \times \;x\left( {x + 1} \right)} \;\;6\left( {x + 1} \right) = 2x\left( {x + 1} \right) + \left( {x - 3} \right)x\\ \Rightarrow 6x + 6 = 2{x^2} + 2x + {x^2} - 3x \Rightarrow 3{x^2} - 7x - 6 = 0\\ \Rightarrow \Delta  = {\left( { - 7} \right)^2} - 4\left( 3 \right)\left( { - 6} \right) = 49 + 72 = 121 \Rightarrow x = \frac{{7 \pm 11}}{6}\\ \Rightarrow \left\{ \begin{array}{l}{x_1} = \frac{{7 - 11}}{6} =  - \frac{4}{6} =  - \frac{2}{3}\\{x_2} = \frac{{7 + 11}}{6} = 3\end{array} \right.\end{array}\)

2) \(\frac{P}{{2 - P}} + \frac{2}{P} = \frac{{ - 3}}{2}\)

\(\begin{array}{l}\frac{P}{{2 - P}} + \frac{2}{P} = \frac{{ - 3}}{2}\;\;\;\mathop  \Rightarrow \limits^{ \times 2P\left( {2 - P} \right)} \;\;2{P^2} + 4\left( {2 - P} \right) =  - 3P\left( {2 - P} \right)\\ \Rightarrow 2{P^2} + 8 - 2P =  - 6P + 3{P^2} \Rightarrow {P^2} - 4P - 8 = 0\\ \Rightarrow \Delta  = {\left( { - 4} \right)^2} - 4\left( 1 \right)\left( { - 8} \right) = 48 \Rightarrow P = \frac{{4 \pm 4\sqrt 3 }}{2}\\ \Rightarrow \left\{ \begin{array}{l}{P_1} = \frac{{4 - 4\sqrt 3 }}{2} = 2 - 2\sqrt 3 \\{P_2} = \frac{{4 + 4\sqrt 3 }}{2} = 2 + 2\sqrt 3 \end{array} \right.\end{array}\)

3) \(\frac{{3y + 5}}{{{y^2} + 5y}} + \frac{{y + 4}}{{y + 5}} = \frac{{y + 1}}{y}\)

\(\begin{array}{l}\frac{{3y + 5}}{{{y^2} + 5y}} + \frac{{y + 4}}{{y + 5}} = \frac{{y + 1}}{y}\;\;\;\mathop  \Rightarrow \limits^{ \times y\left( {y + 5} \right)} \;\;\\\;\left( {3y + 5} \right) + \left( {y + 4} \right)y = \left( {y + 1} \right)\left( {y + 5} \right)\\ \Rightarrow 3y + 5 + {y^2} + 4y = {y^2} + 6y + 5\end{array}\)

\(\Rightarrow\) y = 0  غ ق ق \(\Rightarrow\) جواب ندارد

4) \(2\sqrt x  = \sqrt {3x + 4}\)

\(2\sqrt x  = \sqrt {3x + 4} \;\;\mathop  \Rightarrow \limits^{{{\left( {} \right)}^2}} \;\;4x = 3x + 4 \Rightarrow x = 4\)

5) \(\frac{{1 - \sqrt x }}{{1 + \sqrt x }} = 1 - x\)

\(\begin{array}{l}\frac{{1 - \sqrt x }}{{1 + \sqrt x }} = 1 - x \Rightarrow \frac{{1 - \sqrt x }}{{1 + \sqrt x }} = \left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)\\ \Rightarrow \frac{{1 - \sqrt x }}{{1 + \sqrt x }} - \left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right) = 0\\ \Rightarrow \left( {1 - \sqrt x } \right)\left\{ {\frac{1}{{1 + \sqrt x }} - \left( {1 + \sqrt x } \right)} \right\} = 0\\ \Rightarrow \left( {1 - \sqrt x } \right)\left\{ {\frac{{1 - {{\left( {1 + \sqrt x } \right)}^2}}}{{1 + \sqrt x }}} \right\} = 0\\ \Rightarrow 1 - \sqrt x  = 0 \Rightarrow 1 = \sqrt x  \Rightarrow \\ \Rightarrow \frac{{1 - {{\left( {1 + \sqrt x } \right)}^2}}}{{1 + \sqrt x }} = 0 \Rightarrow 1 - {\left( {1 + \sqrt x } \right)^2} = 0\\ \Rightarrow 1 - \left( {1 + 2\sqrt x  + x} \right) =  - \left( {2\sqrt x  + x} \right) = 0\\ \Rightarrow \sqrt x \left( {2 + \sqrt x } \right) = 0 \Rightarrow \left\{ \begin{array}{l}2 + \sqrt x  = 0 \Rightarrow \sqrt x  =  - 2\;\; \otimes \\\sqrt x  = 0 \Rightarrow \end{array} \right.\end{array}\)

6) \(\frac{5}{{\sqrt x  + 2}} = 2 - \frac{1}{{\sqrt x  - 2}}\)

\(\begin{array}{l}\frac{5}{{\sqrt x  + 2}} = 2 - \frac{1}{{\sqrt x  - 2}}\;\mathop  \Rightarrow \limits^{ \times \left( {x - 4} \right)} \\\;5\left( {\sqrt x  - 2} \right) = 2\left( {x - 4} \right) - \left( {\sqrt x  + 2} \right)\\ \Rightarrow 5\sqrt x  - 10 = 2x - 8 - \sqrt x  - 2\\ \Rightarrow 6\sqrt x  = 2x\mathop {\; \Rightarrow }\limits^{{{\left( {} \right)}^2}} \;36x = 4{x^2}\\ \Rightarrow 4{x^2} - 36x = 0 \Rightarrow 4x\left( {x - 9} \right) = 0\\ \Rightarrow \left\{ \begin{array}{l}4x = 0 \Rightarrow {x_1} = 0\\x - 9 = 0 \Rightarrow {x_2} = 9\end{array} \right.\end{array}\)

7) \(\sqrt {x + 3}  + \sqrt {3x + 1}  = 4\)

\(\begin{array}{l}\sqrt {x + 3} + \sqrt {3x + 1} = 4 \Rightarrow \sqrt {3x + 1} = 4 - \sqrt {x + 3} \\\\{(\,\,)^2} \Rightarrow 3x + 1 = {(4 - \sqrt {x + 3} )^2}\\\\ \Rightarrow 3x + 1 = 16 - 8\sqrt {x + 3} + (x + 3)\\\\ \Rightarrow 8\sqrt {x + 3} = - 2x + 18\\\\{(\,\,)^2} \Rightarrow 64(x + 3) = {( - 2x + 18)^2}\\\\ \Rightarrow 64x + 192 = 4{x^2} - 72x + 324\\\\ \Rightarrow 4{x^2} - 136x + 132 = 0 \Rightarrow {x^2} - 34x + 33 = 0\\\\ \Rightarrow (x - 1)(x - 33) = 0\\\\ \Rightarrow \left\{ \begin{array}{l}x - 1 = 0 \Rightarrow {x_1} = 1\\\\x - 33 = 0 \Rightarrow {x_2} = 33\,\,\,\, \times \end{array} \right. \Rightarrow x = 1\end{array}\)

8)

x: قیمت هر اسباب بازی برحسب هزار تومان

y: تعداد اسباب بازی ها

\(\begin{array}{l}xy = 120 \Rightarrow y = \frac{{120}}{x}\\\left( {x - 1} \right)\left( {y + 4} \right) = 120\\ \Rightarrow xy - y + 4x - 4 = 120\\ \Rightarrow 120 - \frac{{120}}{x} + 4x - 4 = 120\\ \Rightarrow \frac{{120}}{x} = 4\left( {x - 1} \right) \Rightarrow \frac{{30}}{x} = x - 1\\ \Rightarrow x\left( {x - 1} \right) = 30 \Rightarrow {x^2} - x - 30 = 0\\ \Rightarrow \left( {x - 6} \right)\left( {x + 5} \right) = 0\\ \Rightarrow \left\{ \begin{array}{l}x - 6 = 0 \Rightarrow {x_1} = 6\\x + 5 = 0 \Rightarrow {x_2} = - 5\;\; \otimes \end{array} \right. \Rightarrow {x_1} = 6\end{array}\)

بنابراین قیمت هر اسباب بازی قبل از تخفیف 6 هزار تومان بوده است.

9)

x: مدت زمان بر حسب ساعت

\(\frac{1}{{x - 15}}\) : A سرعت کار ماشین

\(\frac{1}{x}\) : B سرعت کار ماشین

\(\begin{array}{l}\frac{1}{{x - 15}} + \frac{1}{x} = \frac{1}{{18}}\;\;\mathop \Rightarrow \limits^{ \times \;18x\left( {x - 15} \right)} \\\;18x + 18\left( {x - 15} \right) = x\left( {x - 15} \right)\\ \Rightarrow 36x - 270 = {x^2} - 15x\\ \Rightarrow {x^2} - 51x + 270 = 0\\ \Rightarrow \left( {x - 6} \right)\left( {x - 45} \right) = 0\\ \Rightarrow \left\{ \begin{array}{l}x - 6 = 0 \Rightarrow {x_1} = 6\;\; \otimes \\x - 45 = 0 \Rightarrow {x_2} = 45\end{array} \right. \Rightarrow {x = 45} \end{array}\)

x - 15 = 30  : A مدت زمان لازم برای ماشین

x = 45  :  B مدت زمان لازم برای ماشین

10)

t₁ : مدت زمان حرکت کشتی در زمان رفت

V₁ : سرعت کشتی در جهت سرعت آب

T₂ : مدت زمان حرکت کشتی در زمان برگشت

V₂ : سرعت کشتی در خلاف جهت سرعت آب

\(\begin{array}{l}\left. \begin{array}{l}{V_1} = {V_2} + 8\\{t_1} = \frac{{144}}{{{V_1}}} = \frac{{144}}{{{V_2} + 8}}\\{t_2} = \frac{{144}}{{{V_2}}}\end{array} \right\}\\ \Rightarrow \left\{ \begin{array}{l}{t_1} + 2 + {t_2} = 17 \Rightarrow {t_1} + {t_2} = 15\\\\ \Rightarrow \frac{{144}}{{{V_2} + 8}} + \frac{{144}}{{{V_2}}} = 15\;\;\mathop \Rightarrow \limits^{ \times \;{V_2}\left( {{V_2} + 8} \right)} \;\\\\144{V_2} + 144\left( {{V_2} + 8} \right) = 15{V_2}\left( {{V_2} + 8} \right)\\\\ \Rightarrow 288{V_2} + 1152 = 15{V_2}^2 + 120{V_2}\\\\ \Rightarrow 15{V_2}^2 - 168{V_2} - 1152 = 0\\\\\Delta = {\left( { - 168} \right)^2} - 4\left( {15} \right)\left( { - 1152} \right) = 97334\\\\ \Rightarrow {V_2} = \frac{{168 \pm 312}}{{30}}\\\\ \Rightarrow \left\{ \begin{array}{l}{V_2} = \frac{{168 - 312}}{{30}} = - \frac{{144}}{{30}}\;\; \otimes \\{V_2} = \frac{{168 + 312}}{{30}} = \frac{{480}}{{30}} = 16\end{array} \right.\\\\ \Rightarrow {V_2} = 16 \Rightarrow {V_1} = 24\end{array} \right.\end{array}\)