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

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

\(\begin{array}{*{20}{l}}\begin{array}{l}\frac{6}{x} = 2 + \frac{{x - 3}}{{x + 1}}\:\:\:\mathop \Rightarrow \limits^{ \times \:x\left( {x + 1} \right)} \:\:\\\\6\left( {x + 1} \right) = 2x\left( {x + 1} \right) + \left( {x - 3} \right)x\end{array}\\\begin{array}{l}\\ \Rightarrow 6x + 6 = 2{x^2} + 2x + {x^2} - 3x\\\\ \Rightarrow 3{x^2} - 7x - 6 = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \Delta = {\left( { - 7} \right)^2} - 4\left( 3 \right)\left( { - 6} \right)\\ = 49 + 72 = 121\\\\ \Rightarrow x = \frac{{7 \pm 11}}{6}\\\end{array}\\{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{x_1} = \frac{{7 - 11}}{6} = - \frac{4}{6} = - \frac{2}{3}}\\{{x_2} = \frac{{7 + 11}}{6} = 3}\end{array}} \right.}\end{array}\)

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

\(\begin{array}{*{20}{l}}\begin{array}{l}\frac{P}{{2 - P}} + \frac{2}{P} = \frac{{ - 3}}{2}\:\:\:\mathop \Rightarrow \limits^{ \times 2P\left( {2 - P} \right)} \:\:\\\\2{P^2} + 4\left( {2 - P} \right) = - 3P\left( {2 - P} \right)\end{array}\\\begin{array}{l}\\ \Rightarrow 2{P^2} + 8 - 2P = - 6P + 3{P^2}\\\\ \Rightarrow {P^2} - 4P - 8 = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \Delta = {\left( { - 4} \right)^2} - 4\left( 1 \right)\left( { - 8} \right) = 48\\\\ \Rightarrow P = \frac{{4 \pm 4\sqrt 3 }}{2}\end{array}\\\begin{array}{l}\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{P_1} = \frac{{4 - 4\sqrt 3 }}{2} = 2 - 2\sqrt 3 }\\{{P_2} = \frac{{4 + 4\sqrt 3 }}{2} = 2 + 2\sqrt 3 }\end{array}} \right.\end{array}\end{array}\)

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 \(2\sqrt x  = \sqrt {3x + 4}\)

\(\begin{array}{l}2\sqrt x = \sqrt {3x + 4} \:\:\mathop \Rightarrow \limits^{{{\left( {} \right)}^2}} \:\:\\\\4x = 3x + 4 \Rightarrow x = 4\end{array}\)

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

\(\begin{array}{*{20}{l}}\begin{array}{l}\frac{{1 - \sqrt x }}{{1 + \sqrt x }} = 1 - x\\\\ \Rightarrow \frac{{1 - \sqrt x }}{{1 + \sqrt x }} = \left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)\end{array}\\\begin{array}{l}\\ \Rightarrow \frac{{1 - \sqrt x }}{{1 + \sqrt x }} - \left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right) = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \left( {1 - \sqrt x } \right)\left\{ {\frac{1}{{1 + \sqrt x }} - \left( {1 + \sqrt x } \right)} \right\} = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \left( {1 - \sqrt x } \right)\left\{ {\frac{{1 - {{\left( {1 + \sqrt x } \right)}^2}}}{{1 + \sqrt x }}} \right\} = 0\end{array}\\\begin{array}{l}\\ \Rightarrow 1 - \sqrt x = 0 \Rightarrow 1 = \sqrt x \Rightarrow {x_1} = 1\end{array}\\\begin{array}{l}\\ \Rightarrow \frac{{1 - {{\left( {1 + \sqrt x } \right)}^2}}}{{1 + \sqrt x }} = 0 \Rightarrow 1 - {\left( {1 + \sqrt x } \right)^2} = 0\end{array}\\\begin{array}{l}\\ \Rightarrow 1 - \left( {1 + 2\sqrt x + x} \right) = - \left( {2\sqrt x + x} \right) = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \sqrt x \left( {2 + \sqrt x } \right) = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{2 + \sqrt x = 0 \Rightarrow \sqrt x = - 2\:\: \otimes }\\{\sqrt x = 0 \Rightarrow {x_2} = 0}\end{array}} \right.\end{array}\end{array}\)

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

\(\begin{array}{*{20}{l}}{\frac{5}{{\sqrt x + 2}} = 2 - \frac{1}{{\sqrt x - 2}}\:\mathop \Rightarrow \limits^{ \times \left( {x - 4} \right)} }\\\begin{array}{l}\:\\5\left( {\sqrt x - 2} \right) = 2\left( {x - 4} \right) - \left( {\sqrt x + 2} \right)\end{array}\\\begin{array}{l}\\ \Rightarrow 5\sqrt x - 10 = 2x - 8 - \sqrt x - 2\end{array}\\\begin{array}{l}\\ \Rightarrow 6\sqrt x = 2x\mathop {\: \Rightarrow }\limits^{{{\left( {} \right)}^2}} \:36x = 4{x^2}\end{array}\\\begin{array}{l}\\ \Rightarrow 4{x^2} - 36x = 0\\\\ \Rightarrow 4x\left( {x - 9} \right) = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{4x = 0 \Rightarrow {x_1} = 0}\\{x - 9 = 0 \Rightarrow {x_2} = 9}\end{array}} \right.\end{array}\end{array}\)

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

\(\begin{array}{*{20}{l}}\begin{array}{l}\sqrt {x + 3} + \sqrt {3x + 1} = 4\\\\ \Rightarrow \sqrt {3x + 1} = 4 - \sqrt {x + 3} \:\:\\\\\mathop \Rightarrow \limits^{{{\left( {} \right)}^2}} \:3x + 1 = {\left( {4 - \sqrt {x + 3} } \right)^2}\end{array}\\\begin{array}{l} \Rightarrow 3x + 1 = 16 - 8\sqrt {x + 3} + \left( {x + 3} \right)\\\\ \Rightarrow 8\sqrt {x + 3} = - 2x + 18\:\\\\\:\mathop \Rightarrow \limits^{{{\left( {} \right)}^2}} \:64\left( {x + 3} \right) = {\left( { - 2x + 18} \right)^2}\end{array}\\\begin{array}{l}\\ \Rightarrow 64x + 192 = 4{x^2} - 72x + 324\\\\ \Rightarrow 4{x^2} - 136x + 132 = 0\\\\ \Rightarrow {x^2} - 34x + 33 = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \left( {x - 1} \right)\left( {x - 33} \right) = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x - 1 = 0 \Rightarrow {x_1} = 1}\\{x - 33 = 0 \Rightarrow {x_2} = 33 \otimes }\end{array}} \right.\\\\ \Rightarrow {x_1} = 1\end{array}\end{array}\)

x: قیمت هر اسباب بازی برحسب هزار تومان

y: تعداد اسباب بازی ها

\(\begin{array}{l}xy = 120 \Rightarrow y = \frac{{120}}{x}\\\\\left( {x - 1} \right)\left( {y + 4} \right) = 120\\\\ \Rightarrow xy - y + 4x - 4 = 120\\\\ \Rightarrow 120 - \frac{{120}}{x} + 4x - 4 = 120\\\\ \Rightarrow \frac{{120}}{x} = 4\left( {x - 1} \right) \Rightarrow \frac{{30}}{x} = x - 1\\\\ \Rightarrow x\left( {x - 1} \right) = 30 \Rightarrow {x^2} - x - 30 = 0\\\\ \Rightarrow \left( {x - 6} \right)\left( {x + 5} \right) = 0\\\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x - 6 = 0 \Rightarrow {x_1} = 6}\\{x + 5 = 0 \Rightarrow {x_2} = - 5\:\: \otimes }\end{array}} \right.\\\\ \Rightarrow {x_1} = 6\end{array}\)

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

x: مدت زمان بر حسب ساعت

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

\(\frac{1}{x}\) : B سرعت کار ماشین

\(\begin{array}{*{20}{l}}{\frac{1}{{x - 15}} + \frac{1}{x} = \frac{1}{{18}}\:\:\mathop \Rightarrow \limits^{ \times \:18x\left( {x - 15} \right)} }\\\begin{array}{l}\\18x + 18\left( {x - 15} \right) = x\left( {x - 15} \right)\end{array}\\\begin{array}{l}\\ \Rightarrow 36x - 270 = {x^2} - 15x\end{array}\\\begin{array}{l}\\ \Rightarrow {x^2} - 51x + 270 = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \left( {x - 6} \right)\left( {x - 45} \right) = 0\end{array}\\\begin{array}{l}\\ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x - 6 = 0 \Rightarrow {x_1} = 6\:\: \otimes }\\{x - 45 = 0 \Rightarrow {x_2} = 45}\end{array}} \right.\\\\ \Rightarrow x = 45\end{array}\end{array}\)

مدت زمان لازم برای ماشین A : 

x - 15 = 30

مدت زمان لازم برای ماشین B : 

x = 45

 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

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