نمونه سوالات تشریحی فصل 7 ریاضی نهم همراه با جواب تشریحی

\(\frac{{2{x^5}{y^6}}}{{ - 4{x^3}{y^2}}} = \frac{2}{{ - 4}}\,\,\,\,\,\,\frac{{{x^5}}}{{{x^3}}}\,\,\,\,\,\frac{{{y^6}}}{{{y^2}}} = - \frac{1}{2}{x^2}{y^4}\)

) گویا هست.

2) گویا هست

3) گویا نیست؛ به علت عبارت \(\sqrt x \)  در صورت کسر

4) گویا نیست؛ زیرا \(\sqrt {{x^2}} = |x|\)

5) گویا نیست؛ به علت وجود عبارت \(|x|\)  در صورت و مخرج

6) گویا هست.

\(\begin{array}{l}1)5x - 10 = 0 \Rightarrow x = \frac{{10}}{5} = 2 \Rightarrow \mathbb{R} - \{ 2\} \\\\2)2{x^2} - 8x + 8 = 0 \Rightarrow 2({x^2} - 4x + 4) = \\\\2{(x - 2)^2} = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2 \Rightarrow \mathbb{R} - \{ 2\} \\\\3)\left\{ \begin{array}{l}x - 1 = 0 \Rightarrow x = 1\\\\x + 1 = 0 \Rightarrow x = - 1\end{array} \right. \Rightarrow \mathbb{R} - \{ - 1,1\} \end{array}\)

\(\begin{array}{l}1)\frac{{{x^3} - 32x}}{{{x^2} + 4x + 4}} = \frac{{8x({x^2} - 4)}}{{{{(x + 2)}^2}}} = \\\\\frac{{8x(x + 2)(x - 2)}}{{{{(x + 2)}^2}}} = \frac{{8x(x - 2)}}{{x + 2}} = \frac{{8{x^2} - 16x}}{{x + 2}}\\\\2)\frac{{5 - 3x}}{{18{x^2} - 60x + 50}} = \frac{{5 - 3x}}{{2(9{x^2} - 30x + 25)}} = \\\\\frac{{ - (3x - 5)}}{{2{{(3x - 5)}^2}}} = \frac{{ - 1}}{{2(3x - 5)}} = \frac{{ - 1}}{{6x - 10}}\\\\3)\frac{{{x^4} - 1}}{{({x^2} + 1)(x + 1)}} = \frac{{({x^2} + 1)({x^2} - 1)}}{{({x^2} + 1)(x + 1)}} = \\\\\frac{{{x^2} - 1}}{{x + 1}} = \frac{{(x + 1)(x - 1)}}{{x + 1}} = x - 1\end{array}\)

\(\begin{array}{l}1)\frac{{3{x^2} + 4{y^2}}}{{{x^2} - {y^2}}} = \frac{{3{{(1)}^2} + 4{{(3)}^2}}}{{{{(1)}^2} - {{(3)}^2}}} = \\\\\frac{{3(1) + 4(9)}}{{(1) - (9)}} = \frac{{3 + 36}}{{ - 8}} = - \frac{{39}}{8}\\\\2)\frac{{4xy + {x^2} - {y^4}}}{{x + {y^2} + 5xy}} = \frac{{4( - 1)(2) + {{( - 1)}^2} - {{(2)}^4}}}{{( - 1) + {{(2)}^2} + 5( - 1)(2)}} = \\\\\frac{{ - 8 + 1 - 16}}{{ - 1 + 4 - 10}} = \frac{{ - 23}}{{ - 7}} = \frac{{23}}{7}\end{array}\)

\(\begin{array}{l}1)\frac{{4{a^2}}}{{2b}} \times \frac{{6{b^2}}}{{15ab}} = \frac{{2{a^2}}}{b} \times \frac{{2{b^2}}}{{5ab}} = \frac{{4{a^2}{b^2}}}{{5{a^2}b}} = \frac{4}{5}b\\\\2)\frac{{3{y^2} - 27}}{{{y^2} - 5y + 6}} \times \frac{{{y^2} + y - 6}}{{3y}} = \\\\\frac{{3(y + 3)(y - 3)}}{{(y - 3)(y - 2)}} \times \frac{{(y + 3)(y - 2)}}{{3y}} = \\\\\frac{{{{(y + 3)}^2}}}{y} = \frac{{{y^2} + 6y + 9}}{y}\end{array}\)

\(\begin{array}{l}1)\frac{{{x^2} - 5x + 6}}{{{x^2} - 4}} \div \frac{{x + 2}}{{{x^2} + 4x + 4}} = \\\\\frac{{{x^2} - 5x + 6}}{{{x^2} - 4}} \times \frac{{{x^2} + 4x + 4}}{{x + 2}} = \\\\\frac{{(x - 3)(x - 2)}}{{(x - 2)(x + 2)}} \times \frac{{{{(x + 2)}^2}}}{{(x + 2)}} = x - 3\\\\2)\frac{{\frac{{{x^2} - 9}}{{{x^2} - 6x + 9}}}}{{\frac{{{x^2} + 6x + 9}}{{{x^2} + 9}}}} = \frac{{({x^2} - 9)({x^2} + 9)}}{{({x^2} - 6x + 9)({x^2} + 6x + 9)}} = \\\\\frac{{(x + 3)(x - 3)({x^2} + 9)}}{{{{(x - 3)}^2}{{(x + 3)}^2}}} = \frac{{{x^2} + 9}}{{(x - 3)(x + 9)}} = \\\\\frac{{{x^2} + 9}}{{{x^2} - 9}}\end{array}\)

\(\begin{array}{l}\frac{{{x^2} - 18x + 56}}{{{x^2} - 16}} \times \cdots = \frac{{x - 14}}{{x + 3}} \Rightarrow \cdots = \\\\\frac{{\frac{{x - 14}}{{x + 3}}}}{{\frac{{{x^2} - 18x + 56}}{{{x^2} - 16}}}} \Rightarrow \cdots = \frac{{\frac{{x - 14}}{{x + 3}}}}{{\frac{{(x - 14)(x - 4)}}{{(x - 4)(x + 4)}}}} = \\\\\frac{{\frac{{x - 14}}{{x + 3}}}}{{\frac{{x - 14}}{{x + 4}}}} = \frac{{(x - 14)(x + 4)}}{{(x + 3)(x - 14)}} = \frac{{x + 4}}{{x + 3}}\end{array}\)

\(\begin{array}{l}1)2x - 3 + \frac{{{x^2} - 5x + 1}}{{4x - 1}} = \\\\\frac{{(2x - 3)(4x - 1) + {x^2} - 5x + 1}}{{4x - 1}} = \\\\\frac{{8{x^2} - 14x + 3 + {x^2} - 5x + 1}}{{4x - 1}} = \frac{{9{x^2} - 19x + 4}}{{4x - 1}}\\\\2)\frac{{a + 4}}{{{a^2} - 4}} - \frac{{a + 6}}{{a + 2}} = \frac{{a + 4 - (a + 6)(a - 2)}}{{{a^2} - 4}} = \\\\\frac{{a + 4 - ({a^2} + 4a - 12)}}{{{a^2} - 4}} = \frac{{a + 4 - {a^2} - 4a + 12}}{{{a^2} - 4}} = \\\\\frac{{ - {a^2} - 3a + 16}}{{{a^2} - 4}}\\\\3)1 - \frac{2}{{{x^2} + 1}} = \frac{{({x^2} + 1) - 2}}{{{x^2} + 1}} = \frac{{{x^2} - 1}}{{{x^2} + 1}}\end{array}\)

\(\begin{array}{l}\frac{{{x^4} - 1}}{{{x^2} + 5x + 6}} = \frac{{({x^2} - 1)({x^2} + 1)}}{{(x + 2)(x + 3)}} = \\\\\frac{{{x^2} - 1}}{{x + 2}} \times \frac{{{x^2} + 1}}{{x + 3}}\end{array}\)

1)

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

\( = \pi {(\frac{{\frac{{x - 1}}{{x + 1}}}}{2})^2} = \pi \frac{{{{(\frac{{x - 1}}{{x + 1}})}^2}}}{4} = \frac{\pi }{4} \times \frac{{{x^2} - 2x + 1}}{{{x^2} + 2x + 1}}\) مساحت دایره

(مساحت دایره – مساحت مربع)\( = \frac{1}{4}\) مساحت قسمت رنگی

\(\begin{array}{l}\frac{1}{4}(\frac{{{x^2} - 2x + 1}}{{{x^2} + 2x + 1}} - \frac{\pi }{4} \times \frac{{{x^2} - 2x + 1}}{{{x^2} + 2x + 1}}) = \\\\\frac{1}{4} \times \frac{{{x^2} - 2x + 1}}{{{x^2} + 2x + 1}}\{ 1 - \frac{\pi }{4}\} = \frac{{4 - \pi }}{{16}} \times \frac{{{x^2} - 2x + 1}}{{{x^2} + 2x + 1}}\end{array}\)

2)

\( = (x + 5) \times \frac{1}{{x - 1}} = \frac{{x + 5}}{{x - 1}}\) مساحت مستطیل بزرگ

\( = (x - 2) \times \frac{1}{{{x^2} - 1}} = \frac{{x - 2}}{{{x^2} - 1}}\) مساحت مستطیل کوچک

مساحت مستطیل کوچک - مساحت مستطیل بزرگ=مساحت قسمت رنگی

\(\begin{array}{l}\frac{{x + 5}}{{x - 1}} - \frac{{x - 2}}{{{x^2} - 1}} = \frac{{(x + 5)(x + 1) - (x - 2)}}{{{x^2} - 1}} = \\\\\frac{{{x^2} + 6x + 5 - x + 2}}{{{x^2} - 1}} = \frac{{{x^2} + 5x + 7}}{{{x^2} - 1}}\end{array}\)

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

\(\begin{array}{l}1)14{x^6}{y^3} \div 7{x^2}y = \frac{{14{x^6}{y^3}}}{{7{x^2}y}} = 2{x^4}{y^2}\\\\2)\frac{{\frac{{x + x + x}}{{y \times y \times y}}}}{{\frac{{x \times x \times x}}{{y + y + y}}}} = \frac{{\frac{{3x}}{{{y^3}}}}}{{\frac{{{x^3}}}{{3y}}}} = \frac{{3x \times 3y}}{{{y^3}{x^3}}} = \\\\\frac{{9xy}}{{{x^3}{y^3}}} = \frac{9}{{{x^2}{y^2}}}\end{array}\)

\(\begin{array}{l}1)\frac{{36x{y^3}{z^4} - 24{x^4}{y^4}{z^5}}}{{4{x^2}y{z^3}}} = \\\\\frac{{36{x^5}{y^3}{z^4}}}{{4{x^2}y{x^3}}} - \frac{{24{x^4}{y^4}{z^5}}}{{4{x^2}y{z^3}}} = \\\\9{x^3}{y^2}z - 6{x^2}{y^3}{z^2}\\\\2)\frac{{3{x^2}{y^4} \times 2{x^5}y - 4x{y^3} \times 3{x^4}{y^2}}}{{6{x^3}{y^2}}} = \\\\\frac{{6{x^7}{y^5} - 12{x^5}{y^5}}}{{6{x^3}{y^2}}} = \frac{{6{x^7}{y^5}}}{{6{x^3}{y^2}}} - \frac{{12{x^5}{y^5}}}{{6{x^3}{y^2}}} = \\\\{x^4}{y^3} - 2{x^2}{y^3}\end{array}\)

\( = {x^2} + x + 1\) خارج قسمت

0=باقی مانده

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

\(\begin{array}{l}ax + b = 3x - \frac{3}{2} \Rightarrow \left\{ \begin{array}{l}a = 3\\\\b = - \frac{3}{2}\end{array} \right.\\\\cx + d = \frac{3}{2}x + \frac{1}{2} \Rightarrow \left\{ \begin{array}{l}c = \frac{3}{2}\\\\d = \frac{1}{2}\end{array} \right.\\\\ \Rightarrow ad - bc = 3 \times \frac{1}{2} + \frac{3}{2} \times \frac{3}{2} = \frac{3}{2} - \frac{9}{4} = \\\\\frac{{6 - 9}}{4} = - \frac{3}{4}\end{array}\)

\(m + \frac{{16}}{9} = 0 \Rightarrow m = - \frac{{16}}{9}\)

گویا

به ازای جمیع مقادیر x

هستند.

\(\frac{{x + 15}}{{x - 4}}\)

یک جمله ای جبری

جمله ی جبری

صفر

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

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

\(\frac{4}{{{x^2} - 1}}\)

نیست

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

\(\frac{{2{x^4} - 1}}{{{x^4} + x}}\)

\(\frac{{ad + bc}}{{bd}}\)

کمتر از درجه

مخرج

یک

صفر

کمتر

باقیمانده – خارج قسمت