نمونه سوالات تستی فصل 7 ریاضی نهم همراه با جواب تشریحی

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

در باقی گزینه ها، عبارت \({x^2} + 1\)  تجزیه ناپذیر می باشد.

عبارت t(n) را به صورت زیر در می آوریم:

\(t(n) = \frac{{an}}{{bn + c}} = \frac{{an}}{{(b + \frac{c}{n})n}} = \frac{a}{{b + \frac{c}{n}}}\)

وقتی n زیاد شود، مقدار \(\frac{c}{n}\)  کوچک تر می شود، در نتیجه مخرج کسر کوچک می شود. در نهایت مقدار t(n) افزایش می یابد.

مخرج گزینه 4 به ازای x=7 صفر می شود.

مخرج عبارت \(\frac{{{x^2} + 1}}{{{x^2} - 3}}\)  را بررسی می کنیم:

\(\begin{array}{l}{x^2} - 3 = 0 \Rightarrow {x^2} = 3 \Rightarrow x = + - \sqrt 3 \\\\ \Rightarrow \{ - \sqrt 3 ,\sqrt 3 \} \end{array}\)

\(\begin{array}{l}a = 1,n = 100 \Rightarrow \frac{{a + 2{a^2} + 3{a^3} + \cdots + n{a^n}}}{{a + {a^2} + {a^3} + \cdots {a^n}}} = \\\\\frac{{1 + 2 + 3 + \cdots + 100}}{{1 + 1 + \cdots + 1}} = \frac{{\frac{{101 \times 100}}{2}}}{{100}} = \frac{{101}}{2} = 50/5\end{array}\)

برای محاسبه \(1 + 2 + 3 + \cdots + 100\)  مثل سال گذشته عمل می کنیم:

\(\begin{array}{l}\frac{\begin{array}{l}1 + 2 + 3 + \cdots + 100 = A\\100 + 99 + \cdots + 2 + 1 = A\end{array}}{{101 + 101 + \cdots + 101 = 2A}}\\\\ \Rightarrow 101 \times 100 = 2A \Rightarrow A = \frac{{101}}{2} \times 100\end{array}\)

\(\begin{array}{l}{x^2} - ax + b = 0 \Rightarrow \left\{ \begin{array}{l}x = 1 \Rightarrow 1 - a + b = 0\\\\x = - 3 \Rightarrow 9 + 3a = b = 0\end{array} \right.\\\\ \Rightarrow \left\{ \begin{array}{l} - a + b = - 1\\\\3a + b = - 9\end{array} \right. \Rightarrow \left\{ \begin{array}{l} - 3a + 3b = - 3\\\\3a + b = - 9\end{array} \right.\\\\ \Rightarrow 4b = - 12 \Rightarrow b = - 3 \Rightarrow - a + b = - 1\\\\ \Rightarrow - a - 3 = - 1 \Rightarrow a = - 2 \Rightarrow a + b = - 5\end{array}\)

\(\begin{array}{l}\frac{{{x^8} - {y^8}}}{{({x^4} + {y^4}){{({x^2} - {y^2})}^2}}} = \\\\\frac{{({x^4} + {y^4})({x^4} - {y^4})}}{{({x^4} + {y^4}){{({x^2} - {y^2})}^2}}} = \frac{{{x^4} - {y^4}}}{{{{({x^2} - {y^2})}^2}}} = \\\\\frac{{({x^2} + {y^2})({x^2} - {y^2})}}{{{{({x^2} - {y^2})}^2}}} = \frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}}\end{array}\)

\(\begin{array}{l}\frac{{2{x^2} - 5}}{{{x^2} + 6x + 8}} - \frac{x}{{x + 2}} = \\\\\frac{{2{x^2} - 5}}{{{x^2} + 6x + 8}} - \frac{{x(x + 4)}}{{{x^2} + 6x + 8}} = \\\\\frac{{2{x^2} - 5 - ({x^2} + 4x)}}{{{x^2} + 6x + 8}} = \frac{{{x^2} - 4x - 5}}{{{x^2} + 6x + 8}}\end{array}\)

\(\begin{array}{l}\frac{{4{x^2} - 4}}{{2x + 2}} \div \frac{2}{{x - 3}} = \\\\\frac{{4(x = 1)(x - 1)}}{{2(x + 1)}} \times \frac{{(x - 3)}}{2} = \\\\(x - 1) \times (x - 3) = {x^2} - 4x + 3\end{array}\)

\(\begin{array}{l}{x^2} + {y^2} = - 2x \Rightarrow {x^2} + 2xy + {y^2} = \\\\0 \Rightarrow {(x + y)^2} = 0 \Rightarrow x + y = 0 \Rightarrow x = - y\\\\\frac{{{{(2x - y)}^4}}}{{{{(2x + y)}^4}}} = {(\frac{{2x - y}}{{2x + y}})^4} = {(\frac{{ - 2y - y}}{{ - 2y + y}})^4} = \\\\{(\frac{{ - 3y}}{{ - y}})^4} = {3^4} = 81\end{array}\)

\(\begin{array}{l}\frac{{(1 + \frac{1}{2})(1 + \frac{1}{3}) \times \cdots \times (1 + \frac{1}{n})}}{{(1 - \frac{1}{2})(1 - \frac{1}{3}) \times \cdots \times (1 - \frac{1}{n})}} = 55\\\\ \Rightarrow \frac{{\frac{3}{2} \times \frac{4}{3} \times \cdots \times \frac{n}{{n - 1}} \times \frac{{n + 1}}{n}}}{{\frac{1}{2} \times \frac{2}{3} \times \cdots \times \frac{{n - 2}}{{n - 1}} \times \frac{{n - 1}}{n}}} = 55\\\\ \Rightarrow \frac{{\frac{1}{2} \times (n + 1)}}{{1 \times \frac{1}{n}}} = 55 \Rightarrow \frac{{n(n + 1)}}{2} = 55\\\\ \Rightarrow {n^2} + n = 110 \Rightarrow {n^2} + n - 110 = 0\\\\ \Rightarrow (n - 10)(n + 1) = 0 \Rightarrow n = 10\end{array}\)

\(\begin{array}{l}\frac{{2x}}{{a{x^2} + ax}} \div \frac{{x - 3}}{{2{x^2} - 4x - 6}} = \frac{1}{6}\\\\ \Rightarrow \frac{{2x}}{{ax(x + 1)}} \div \frac{{x - 3}}{{2(x - 3)(x + 1)}} = \frac{1}{6}\\\\ \Rightarrow \frac{2}{{a(x + 1)}} \div \frac{1}{{2(x + 1)}} = \frac{1}{6}\\\\ \Rightarrow \frac{2}{{a(x + 1)}} \times \frac{{2(x + 1)}}{1} = \frac{1}{6} \Rightarrow \frac{4}{a} = \frac{1}{6} \Rightarrow a = 24\end{array}\)

می دانیم که \(\frac{{x - y}}{{xy}} = \frac{1}{y} - \frac{1}{x}\) ؛ بنابراین داریم:

\(\begin{array}{l}\left. \begin{array}{l}\frac{1}{{x(x + 1)}} = \frac{1}{x} - \frac{1}{{x + 1}}\\\\\frac{1}{{(x + 1)(x + 2)}} = \frac{1}{{x + 1}} - \frac{1}{{x + 2}}\\\\\frac{1}{{(x + 2)(x + 3)}} = \frac{1}{{x + 2}} - \frac{1}{{x + 3}}\\\\\frac{1}{{(x + 99)(x + 100)}} = \frac{1}{{x + 99}} - \frac{1}{{x + 100}}\end{array} \right\}\\\\ \Rightarrow \frac{1}{x} - \frac{1}{{x + 1}} + \frac{1}{{x + 1}} - \frac{1}{{x + 2}} + \frac{1}{{x + 2}} - \frac{1}{{x - 3}}\\\\ + \cdots + \frac{1}{{x + 99}} - \frac{1}{{x + 100}} = \frac{1}{{200}}\\\\ \Rightarrow \frac{1}{x} - \frac{1}{{x + 100}} = \frac{1}{{200}}\\\\\frac{{100}}{{x(x + 100)}} = \frac{1}{{200}} \Rightarrow {x^2} + 100x = 20000\\\\ \Rightarrow {x^2} + 100x - 20000 = 0\\\\ \Rightarrow (x - 100)(x + 200) = 0 \Rightarrow \left\{ \begin{array}{l}x = 100\\\\x = - 200\end{array} \right.\\\\ \Rightarrow x = 100\end{array}\)

\(\begin{array}{l}x - 200 \Rightarrow x = 2\\\\{x^5} - 3x + 1 = {2^5} - 3(2) + 1 = 32 - 6 + 1 = 27\end{array}\)

\(\begin{array}{l}x + 1 = 0 \Rightarrow x = - 1\\\\ \Rightarrow m{( - 1)^2} + 3( - 1) + n = - 4 \Rightarrow m - 3 + n = - 4\\\\ \Rightarrow m + n = - 1\\\\x - 2 = 0 \Rightarrow x = 2 \Rightarrow m{(2)^2} + 3(2) + n = 11\\\\ \Rightarrow 4m + 6 + n = 11 \Rightarrow 4m + n = 5\\\\ \Rightarrow \left\{ \begin{array}{l}m + n = - 1\\\\4m + n = 5\end{array} \right. \Rightarrow \left\{ \begin{array}{l} - m - n = 1\\\\4m + n = 5\end{array} \right.\\\\ \Rightarrow 3m = 6 \Rightarrow m = 2 \Rightarrow n = - 3\\\\ \Rightarrow 2m + 3n = - 5\end{array}\)

\(\begin{array}{l}\frac{{{x^4} - {y^4}}}{{x - y}} = \frac{{({x^2} - {y^2})({x^2} + {y^2})}}{{x - y}} = \\\\\frac{{(x - y)(x + y)({x^2} + {y^2})}}{{x - y}} = \\\\(x + y)({x^2} + {y^2}) = {x^3} + {x^2}y + x{y^2} + {y^3}\end{array}\)

\(\begin{array}{l}x + 2 = 0 \Rightarrow x = - 2 \Rightarrow a{( - 2)^2} + 3( - 2) - 1 = - 1\\\\ \Rightarrow 4a - 6 - 1 = - 1 \Rightarrow 4a = 6\\\\ \Rightarrow a = 1/5\end{array}\)

\(\begin{array}{l}x + 1 = 0 \Rightarrow x = - 1 \Rightarrow p( - 1) = (x + 1)Q + 3\\\\x - 2 = 0 \Rightarrow x = 2 \Rightarrow p(2) = (x - 2)Q' + 6\end{array}\)

می دانیم باقی مانده p(x) بر \({x^2} - x - 2\)  از درجه یک به صورت ax+b می باشد و همچنین می دانیم که \({x^2} - x - 2 = (x + 1)(x - 2)\) ؛ بنابراین:

\(\begin{array}{l}p(x) = (x + 1)(x - 2){Q^7} + ax + b\\\\ \Rightarrow \left\{ \begin{array}{l}x = - 1 \Rightarrow p( - 1) = 3 = - a + b\\\\x = 2 \Rightarrow p(2) = 6 = 2a + b\end{array} \right.\\\\ \Rightarrow \left\{ \begin{array}{l} - a + b = 3\\\\2a + b = 6\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a - b = - 3\\\\2a + b = 6\end{array} \right.\\\\ \Rightarrow 3a = 3 \Rightarrow a = 1 \Rightarrow b = 4\\\\ \Rightarrow ax + b = x + 4\end{array}\)

\(\begin{array}{l}x + 1 = 0 \Rightarrow x = - 1\\\\ \Rightarrow \left\{ \begin{array}{l}2{x^5} + ax + 7 = 2{( - 1)^5} + a( - 1) + 7 = \\ - 2 - a + 7 = - a + 5\\\\3{x^3} - 6{x^2} + 5 = 3{( - 1)^3} - 6{( - 1)^2} + 5 = \\ - 3 - 6 + 5 = - 4\end{array} \right.\\\\ \Rightarrow - a + 5 = - 4 \Rightarrow - a = - 9 \Rightarrow a = 9\end{array}\)

\(\begin{array}{l}\frac{{{x^{ - 2}} - {y^{ - 2}}}}{{{x^{ - 4}} - {y^{ - 4}}}} = \frac{{{{(\frac{1}{x})}^2} - {{(\frac{1}{y})}^4}}}{{{{(\frac{1}{x})}^4} - {{(\frac{1}{y})}^4}}} = \\\\\frac{{{{(\frac{1}{x})}^2} - {{(\frac{1}{y})}^2}}}{{[{{(\frac{1}{x})}^2} - {{(\frac{1}{y})}^2}][{{(\frac{1}{x})}^2} + {{(\frac{1}{y})}^2}]}} = \frac{1}{{\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}}}} = \\\\\frac{1}{{\frac{{{x^2} + {y^2}}}{{{x^2}{y^2}}}}} = \frac{{{x^2}{y^2}}}{{{x^2} + {y^2}}}\end{array}\)