جواب کاردرکلاس صفحه ۳۵ ریاضی دهم انسانی

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جواب صفحه ۳۵ ریاضی دهم انسانی

معادله های زیر را حل کنید.

\(۱ + \frac{8}{{{x^2}}} = \frac{4}{x}\) الف

\(\begin{array}{l}1 + \frac{8}{{{x^2}}} = \frac{4}{x} \Rightarrow 1 + \frac{8}{{{x^2}}} – \frac{4}{x} = 0 \Rightarrow \frac{{{x^2}}}{{{x^2}}} + \frac{8}{{{x^2}}} – \frac{{4x}}{{{x^2}}} = 0\\\\ \Rightarrow \frac{{{x^2} – 4x + 8}}{{{x^2}}} = 0 \Rightarrow {x^2} – 4x + 8 = 0\\\\ \Rightarrow \Delta = {b^2} – 4ac = {( – 4)^2} – 4(1)(8) = 16 – 32 = – 16 < 0\end{array}\)

معادله ریشه ندارد.

\(\frac{{x – 2}}{{x – 4}} = \frac{{x + 1}}{{x + 3}}\) ب

\(\begin{array}{l}\frac{{x – 2}}{{x – 4}} = \frac{{x + 1}}{{x + 3}} \Rightarrow \frac{{x – 2}}{{x – 4}} – \frac{{x + 1}}{{x + 3}} = 0\\\\ \Rightarrow \frac{{(x – 2)(x + 3)}}{{(x – 4)(x + 3)}} – \frac{{(x + 1)(x – 4)}}{{(x + 3)(x – 4)}} = 0\\\\ \Rightarrow \frac{{{x^2} + x – 6 – {x^2} + 3x + 4}}{{(x – 3)(x – 4)}} = 0 \Rightarrow \frac{{4x – 2}}{{(x – 3)(x – 4)}} = 0\\\\ \Rightarrow 4x – 2 = 0 \Rightarrow x = \frac{1}{2}\end{array}\)

\(\frac{{24}}{{10 + m}} + 1 = \frac{{24}}{{10 – m}}\) پ

\(\begin{array}{l}\frac{{24}}{{10 + m}} + 1 = \frac{{24}}{{10 – m}} \Rightarrow \frac{{24}}{{10 + m}} + 1 – \frac{{24}}{{10 – m}} = 0 \Rightarrow \\\\\frac{{24(10 – m)}}{{(10 + m)(10 – m)}} + \frac{{(10 + m)(10 – m)}}{{(10 + m)(10 – m)}} – \frac{{24(10 – m)}}{{(10 + m)(10 – m)}} = 0\\\\\frac{{240 – 24m + 100 – {m^2} – 240 – 24m}}{{(10 + m)(10 – m)}} = 0\\\\ \Rightarrow – {m^2} – 48m + 100 = 0 \Rightarrow {m^2} + 48m – 100 = 0\\\\ \Rightarrow (m + 50)(m – 2) = 0 \Rightarrow \left\{ \begin{array}{l}m = – 50\\\\m = 2\end{array} \right.\end{array}\)

\(\frac{{y + 2}}{{y + 3}} – \frac{{{y^2}}}{{{y^2} – 9}} = 1 – \frac{{y – 1}}{{3 – y}}\) ت

\(\begin{array}{l}\frac{{y + 2}}{{y + 3}} – \frac{{{y^2}}}{{{y^2} – 9}} = 1 – \frac{{y – 1}}{{3 – y}}\\\\ \Rightarrow \frac{{y + 2}}{{y + 3}} – \frac{{{y^2}}}{{(y – 3)(y + 3)}} – 1 + \frac{{y – 1}}{{ – (3 – y)}} = 0\\\\ \Rightarrow \frac{{(y + 2)(y – 3)}}{{(y + 3)(y – 3)}} – \frac{{{y^2}}}{{(y – 3)(y + 3)}} – \frac{{(y – 3)(y + 3)}}{{(y – 3)(y + 3)}} – \\\\\frac{{(y – 1)(y + 3)}}{{(y – 3)(y + 3)}} = 0\\\\ \Rightarrow \frac{{{y^2} – y – 6 – {y^2} – {y^2} + 9 – {y^2} – 2y + 3}}{{(y – 3)(y + 3)}} = 0\\\\ \Rightarrow – 2{y^2} – 3y + 6 = 0 \Rightarrow 2{y^2} + 3y – 6 = 0\\\\ \Rightarrow \Delta = {3^2} – 4(2)( – 6) = 9 + 48 = 57 > 0\\\\ \Rightarrow y = \frac{{ – 3 \pm \sqrt {57} }}{4} \Rightarrow \left\{ \begin{array}{l}y = \frac{{ – 3 – \sqrt {57} }}{4}\\\\y = \frac{{ – 3 + \sqrt {57} }}{4}\end{array} \right.\end{array}\)

ت به ازای چه مقدار a، معادله \(\frac{x}{{a – x}} + \frac{{a – x}}{x} = \frac{a}{x}\) دارای جواب x=2 است؟

\(\begin{array}{l}\left\{ \begin{array}{l}\frac{x}{{a – x}} + \frac{{a – x}}{x} = \frac{a}{x}\\\\x = 2\end{array} \right. \Rightarrow \frac{2}{{a – 2}} + \frac{{a – 2}}{2} = \frac{a}{2}\\\\ \Rightarrow \frac{2}{{a – 2}} + \frac{{a – 2}}{2} – \frac{a}{2} = 0 \Rightarrow \frac{{2(2)}}{{2(a – 2)}} + \frac{{(a – 2)(a – 2)}}{{2(a – 2)}} – \\\\\frac{{a(a – 2)}}{{2(a – 2)}} = 0 \Rightarrow \frac{{4 + {a^2} + 4 – 4a – {a^2} + 2a}}{{2(a – 2)}} = 0\\\\ \Rightarrow – 2a + 8 = 0 \Rightarrow a = 4\end{array}\)