گام به گام تمرین صفحه 37 درس معادلۀ درجۀ دوم ریاضی و آمار (1)

\(\begin{array}{l}\frac{{3x - 5}}{{x + 3}} = 1 \Rightarrow \frac{{3x - 5}}{{x + 3}} - 1 = 0 \Rightarrow \frac{{3x - 5}}{{x + 3}} - \frac{{x + 3}}{{x + 3}} = 0\\\\ \Rightarrow \frac{{3x - 5 - x - 3}}{{x + 3}} = 0 \Rightarrow 2x - 2 = 0 \Rightarrow x = 1\end{array}\)

\(\begin{array}{l}\frac{{3x - 2}}{x} + \frac{{2x + 5}}{{x + 3}} = 5 \Rightarrow \frac{{3x - 2}}{x} + \frac{{2x + 5}}{{x + 3}} - 5 = 0\\\\ \Rightarrow \frac{{(3x - 2)(x + 3)}}{{x(x + 3)}} + \frac{{(2x + 5)x}}{{x(x + 3)}} - \frac{{5x(x + 3)}}{{x(x + 3)}} = 0\\\\ \Rightarrow \frac{{3{x^2} + 9x - 2x - 6 + 2{x^2} + 5x - 5{x^2} - 15x}}{{x(x + 3)}} = 0\\\\ \Rightarrow \frac{{ - 3x - 6}}{{x(x + 3)}} = 0 \Rightarrow - 3x - 6 = 0 \Rightarrow x = - 2\end{array}\)

\(\begin{array}{l}\frac{2}{{x + 2}} + \frac{x}{{x + 2}} = x + 3 \Rightarrow \frac{{2 + x}}{{x + 2}} - x - 3 = 0\\\\ \Rightarrow 1 - x - 3 = 0 \Rightarrow x = - 2\end{array}\)

جواب x=-2 غیر قابل قبول است؛ زیرا مخرج را صفر می کند، در نتیجه معادله جواب ندارد.

\(\begin{array}{l}\frac{{{x^2} - 2x + 2}}{{{x^2} - 2x}} - \frac{{1 + x}}{x} = \frac{{x - 1}}{{x - 2}}\\\\ \Rightarrow \frac{{{x^2} - 2x + 2}}{{{x^2} - 2x}} - \frac{{1 + x}}{x} - \frac{{x - 1}}{{x - 2}} = 0\\\\ \Rightarrow \frac{{{x^2} - 2x + 2}}{{x(x - 2)}} - \frac{{(x + 1)(x - 2)}}{{x(x - 2)}} - \frac{{x(x - 1)}}{{x(x - 2)}} = 0\\\\ \Rightarrow \frac{{{x^2} - 2x + 2}}{{x(x - 2)}} - \frac{{{x^2} - x - 2}}{{x(x - 2)}} - \frac{{{x^2} - x}}{{x(x - 2)}} = 0\\\\ \Rightarrow \frac{{{x^2} - 2x + 2 - {x^2} + x + 2 - {x^2} + x}}{{x(x - 2)}} = \frac{{ - {x^2} + 4}}{{x(x - 2)}} = 0\\\\ \Rightarrow - {x^2} + 4 = 0 \Rightarrow {x^2} = 4 \Rightarrow x = - 2\end{array}\)

x=2 غیر قابل قبول است؛ زیرا مخرج را صفر می کند.

\(\begin{array}{l}\frac{3}{{x - 1}} - \frac{2}{{x + 3}} = \frac{4}{{x - 2}} \Rightarrow \frac{3}{{x - 1}} - \frac{2}{{x + 3}} - \frac{4}{{x - 2}} = 0\\\\\frac{{3(x - 2)(x + 3)}}{{(x - 1)(x - 2)(x + 3)}} - \frac{{2(x - 1)(x - 2)}}{{(x - 1)(x - 2)(x + 3)}} - \\\\\frac{{4(x - 1)(x + 3)}}{{(x - 1)(x - 2)(x + 3)}} = 0\\\\ \Rightarrow \frac{{3{x^2} + 3x - 18 - 2{x^2} + 6x - 4 - 4{x^2} - 8x + 12}}{{(x - 1)(x - 2)(x + 3)}} = 0\\\\ \Rightarrow - 3{x^2} + x - 10 = 0 \Rightarrow \Delta = {1^2} - 4( - 3)( - 10) = - 119 < 0\end{array}\)

معادله، جوابی ندارد.

\(\begin{array}{l}\frac{{11}}{{{x^2} - 4}} + \frac{{x + 3}}{{2 - x}} = \frac{{2x - 3}}{{x + 2}} \Rightarrow \frac{{11}}{{{x^2} - 4}} + \frac{{x + 3}}{{2 - x}} - \frac{{2x - 3}}{{x + 2}} = 0\\\\ \Rightarrow \frac{{11}}{{{x^2} - 4}} + \frac{{x + 3}}{{ - (x - 2)}} - \frac{{2x - 3}}{{x + 2}} = 0\\\\ \Rightarrow \frac{{11}}{{(x - 2)(x + 2)}} - \frac{{(x + 3)(x + 2)}}{{(x - 2)(x + 2)}} - \frac{{(2x - 3)(x - 2)}}{{(x - 2)(x + 2)}} = 0\\\\ \Rightarrow \frac{{11 - {x^2} - 5x - 6 - 2{x^2} + 4x + 3x - 6}}{{(x - 2)(x + 2)}} = 0\\\\ \Rightarrow - 3{x^2} + 2x - 1 = 0 \Rightarrow \Delta = {2^2} - 4( - 3)( - 1) = - 8 < 0\end{array}\)

معادله، جوابی ندارد.

\(\begin{array}{l}\frac{1}{{2k}} + \frac{1}{{2k + 2}} = \frac{5}{{12}} \Rightarrow \frac{1}{{2k}} + \frac{1}{{2k + 2}} - \frac{5}{{12}} = 0\\\\ \Rightarrow \frac{{6(k + 1)}}{{12k(k + 1)}} + \frac{{6k}}{{12k(k + 1)}} - \frac{{5k(k + 1)}}{{12k(k + 1)}} = 0\\\\ \Rightarrow \frac{{6k + 6 + 6k - 5{k^2} - 5k}}{{12k(k + 1)}} = 0 \Rightarrow - 5{k^2} + 7k + 6 = 0\\\\ \Rightarrow \Delta = {7^2} - 4( - 5)(6) = 49 + 120 = 169\\\\k = \frac{{ - 7 \pm \sqrt {169} }}{{ - 10}} = \frac{{ - 7 \pm 13}}{{ - 10}} \Rightarrow k = \frac{{ - 7 - 13}}{{ - 10}} = 2\end{array}\)

آن دو عدد، 4 و 6 می باشند.

\(\begin{array}{l}\frac{1}{x} + \frac{1}{{x + 3}} = \frac{1}{4} \Rightarrow \frac{1}{x} + \frac{1}{{x + 3}} - \frac{1}{4} = 0\\\\ \Rightarrow \frac{{4(x + 3)}}{{4x(x + 3)}} + \frac{{4x}}{{4x(x + 3)}} - \frac{{x(x + 3)}}{{4x(x + 3)}} = 0\\\\ \Rightarrow \frac{{4x + 12 + 4x - {x^2} - 3x}}{{4x(x + 3)}} = 0\\\\ \Rightarrow - {x^2} + 5x + 12 = 0 \Rightarrow \Delta = {5^2} - 4( - 1)(12) = 73 > 0\\\\ \Rightarrow x = \frac{{ - 5 \pm \sqrt {73} }}{{ - 2}} \Rightarrow x > 0 \Rightarrow x = \frac{{ - 5 - \sqrt {73} }}{{ - 2}} = \frac{{5 + \sqrt {73} }}{2}\end{array}\)

\(\begin{array}{l}t = - 3 \Rightarrow \frac{{4 - ( - 3)}}{{2 - 2( - 3)}} = \frac{{3(9) + k}}{{100 - 68}} \Rightarrow \frac{7}{8} = \frac{{27 + k}}{{32}}\\\\ \Rightarrow \frac{{7 \times 32}}{8} = (27 + k) \Rightarrow k = 28 - 27 \Rightarrow k = 1\end{array}\)