جواب تمرین صفحه 85 درس 5 ریاضی نهم (عبارت های جبری)

الف

\(\begin{array}{l}25{m^2}( - 8{m^3}) - \frac{1}{4}{m^2}( - 8{m^3}) = \\\\ - 200{m^5} + 2{m^5} = 198{m^5}\end{array}\)

ب 

\(\begin{array}{l}7{a^3} - 4{b^3} + 5{c^3} - {a^3} + 9{b^3} + 11{c^3} = \\\\6{a^3} + 5{b^3} + 16{c^3}\end{array}\)

ج \({x^{2m}} - 2{x^m} + 1\)

د \(x - \left[ {y - x - y + 1} \right] = x + x - 1 = 2x - 1\)

الف ارتفاع ذوزنقه:

\(\begin{array}{l}{h^2} = 9{m^4}{n^2} - {m^4}{n^2} = 8{m^4}{n^2}\\\\h = {m^2}n\sqrt 8 \end{array}\)

محیط:

\(3{m^2}n + 2{m^2}n + 5{m^2}n + 3{m^2}n + 3{m^2}n + 2{m^2}n = 18{m^2}n\)

مساحت:

\(3{m^2}n \times 2{m^2}n + \frac{{{m^2}n\sqrt 8 (3{m^2}n + 5{m^2}n)}}{2} = 6{m^4}{n^2} + 4{m^4}{n^2}\sqrt 8 \)

ب ابتدا با استفاده از رابطه فیثاغورس ضلع مثلث را محاسبه می کنیم:

\(25{x^2}{y^3} = 9{x^2}{y^2} + (\) ضلع مثلث\({)^2}\)

(ضلع مثلث\({)^2} = 25{x^2}{y^3} - 9{x^2}{y^2} = 16{x^2}{y^2}\)

\( = 4xy\)ضلع مثلث

محیط \( = 3xy + (2 \times 4xy) + 4xy + 5xy = 20xy\)

مساحت  \( = (3xy \times 4xy) + (3xy \times 4xy \times \frac{1}{2}) = 4{x^2}{y^2} + 6{x^2}{y^2} = 18{x^2}{y^2}\)

الف \(25{y^2} - 30xy + 9{x^2}\)

ب \(9{a^4} + 6{a^3} + {a^2}\)

ج \(64{x^2} - \frac{{16}}{3}x + \frac{1}{9}\)

د \({(2/7 + 3/3)^2} = {6^2} = 36\)

الف \({x^2} + 2xy + {y^2} - {x^2} + 2xy - {y^2} = 4xy\)

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

الف \(2{x^3} + 8{x^2} + 8x = 2x({x^2} + 2x + 4)\)

ب \(3{a^3}b - 12a{b^3} + {a^3}{b^3} = ab(3{a^2} - 12{b^2} + {a^2}{b^2})\)

ج \(a(x + 1) + b{(x + 1)^2} = (x + 1) + (a + b(x + 1))\)

د \({a^3} - 2{a^2} + a = a({a^2} - 2a + 1) = a{(a - 1)^2}\)

هـ \({x^2}{y^2} - 4xy + 4 = {(xy - 2)^2}\)

و \(25{x^4}30{x^3} + 9{x^2} = {x^2}(25{x^2} + 30x + 9)\)

\(\begin{array}{l}{(a + ( - b))^2} = {a^2} + 2a( - b) + {( - b)^2} = \\\\{a^2} - 2ab + {b^2}\end{array}\)