جد معامل x ² في مفكوك ${(x^3+\frac{2}{x^2})}^9$

نفرض الحد الذي يحوي هوي على x² هو r

$\begin{array}{rl}& \mathrm{Pr}=C_{r-1}^n{x}^{n-r+1}{y}^{r-1}\\ & \mathrm{Pr}=C_{r-1}^9{\left(x^3\right)}^{9-r+1}{\left(\frac{2}{x^2}\right)}^{r-1}\\ & \mathrm{Pr}=C_{r-1}^9{\left(x^3\right)}^{10-r}(2{)}^{r-1}{\left(x^{-2}\right)}^{r-1}\\ & Pr=C_{r-1}^9{x}^{30-3r}(2{)}^{r-1}{x}^{-2r+2}\\ & x^{30-3r}\cdot x^{-2+2r}=x^2=x^{32-5r}=x^2\\ & 32-5r=2⟹5r=30]÷5⟹r=6\\ & P6=C_{6-1}^9{x}^{30-18}(2{)}^{6-1}{x}^{-12+2}\\ & P6=\frac{9\times 8\times 7\times 6\times 5}{5\times 4\times 3\times 2\times 1}{x}^{12}(2{)}^5{x}^{-10}=3\times 2\times 7\times 3\times 32x^2=4032x^2\end{array}$

تمارين (4-1)

1- جد مفكوك كل مما يأتي:

- $(3a-b{)}^{4^4}$

$\begin{array}{rl}& =C_0^4(3a{)}^4-C_1^4(3a{)}^3\left(b\right)+C_2^4(3a{)}^2\cdot (b{)}^2-C_3^4\left(3a\right)\left(b^3\right)+C_4^4{b}^4\\ & =81a^4-4\left(27a^3\right)\left(b\right)+\frac{4\times 3}{2\times 1}\left(9a^2\right)\left(b^2\right)-4\left(3a\right)\left(b^3\right)+b^4\\ & =81a^4-108a^3\cdot b+54a^2\cdot b^2-12a{b}^3+b^4\end{array}$

- ${(3x^2+2y)}^3$

$\begin{array}{rl}& =C_0^3{\left(3x^2\right)}^3+C_1^3{\left(3x^2\right)}^2\left(2y\right)+C_2^3\left(3x^2\right)(2y{)}^2+C_3^3(2y{)}^3\\ & 27x^6+3\left(9\right)x^4\left(2y\right)+3\left(3x^2\right)\left(4y^2\right)+8y^3=27x^6+54x^{4}y+36x^2{y}^2+8y^3\end{array}$

- ${(2x-\frac{1}{2x})}^6$

$\begin{array}{rl}=& C_0^6(2x{)}^6-C_1^6(2x{)}^5\left(\frac{1}{2x}\right)+C_2^6(2x{)}^4{\left(\frac{1}{2x}\right)}^2-C_3^6(2x{)}^3{\left(\frac{1}{2x}\right)}^3+C_4^6(2x{)}^2{\left(\frac{1}{2x}\right)}^4-C_5^6(2x){\left(\frac{1}{2x}\right)}^5+C_6^6{\left(\frac{1}{2x}\right)}^6\\ =& 64x^6-6\left(32x^5\right)\left(\frac{1}{2x}\right)+\frac{6\times 5}{2\times 1}\left(16x^4\right)\left(\frac{1}{4x^2}\right)-\frac{6\times 5\times 4}{3\times 2\times 1}\left(8x^3\right)\left(\frac{1}{8x^3}\right)+\frac{6\times 5\times 4\times 3}{4\times 3\times 2\times 1}\left(4x^2\right)\left(\frac{1}{16x^4}\right)-\\ & 6\left(2x\right)\left(\frac{1}{32x^5}\right)+\frac{1}{64x^6}\\ =& 64x^6-96x^4+60x^2-20+\frac{15}{4x^2}-\frac{3}{8x^4}+\frac{1}{64x^6}\end{array}$

2- جد الحد الثالث في مفكوك $(x-3y{)}^7$

$\begin{array}{rl}& Pr=C_{r-1}^n{x}^{n-r+1}{y}^{r-1}\\ & P3=C_{3-1}^7{x}^{7-3+1}(-3y{)}^{3-1}\\ & P3=\frac{7\times 6}{2\times 1}\cdot x^5(-3y{)}^2=21x^5\left(9y^2\right)=189x^5{y}^2\end{array}$

3- جد الحد السادس في مفكوك ${(\frac{x^2}{2}-\frac{x^2}{3})}^8$

$\begin{array}{rl}& Pr=C_{r-1}^n{x}^{n-r+1}{y}^{r-1}\\ & P6=C_{6-1}^8{\left(\frac{x^2}{2}\right)}^{8-6+1}{(-\frac{x}{3})}^{6-1}\Rightarrow P6=\frac{8\times 7\times 6\times 5\times 4}{5\times 4\times 3\times 2\times 1}{\left(\frac{x^2}{2}\right)}^3{\left(\frac{-x}{3}\right)}^5=56\times \frac{x^6}{8}\times \frac{-x^5}{243}=\frac{-7x^{11}}{243}\end{array}$

4- جد الحد الأوسط في مفكوك ${(a-\frac{2}{a})}^{12}$

$\begin{array}{rl}& r=\frac{n}{2}+1=\frac{12}{2}+1=7\\ & Pr=C_{r-1}^n{x}^{n-r+1}{y}^{r-1}\\ & P7=C_{7-1}^{12}{a}^{12-7+1}{\left(\frac{-2}{a}\right)}^6=\frac{12\times 11\times 10\times 9\times 8\times 7\times 6}{6\times 5\times 4\times 3\times 2\times 1}{a}^6\frac{64}{a^6}\\ & =2\times 11\times 2\times 3\times 7\times 8\times 64=59136\end{array}$

5- جد الحد الأوسط في مفكوك ${(\frac{x}{2}-3)}^8$

$\begin{array}{rl}& r1=\frac{n+1}{2}=\frac{7+1}{2}=4 , r2=\frac{n+1}{2}+1=\frac{7+1}{2}+1=5\\ & Pr=C_{r-1}^n{x}^{n-r+1}{y}^{r-1}\\ & P4=C_{4-1}^7\left(2a{)}^{7-4+1}\right(-1{)}^{4-1}=\frac{7\times 6\times 5}{3\times 2\times 1}\left(2a{)}^4\right(-1{)}^3=35\left(16\right)a^4(-1)=-560a^4\\ & P5=C_{5-1}^7\left(2a{)}^{7-5+1}\right(-1{)}^{5-1}=\frac{7\times 6\times 5\times 4}{4\times 3\times 2\times 1}\left(2a{)}^3\right(-1{)}^4=35\left(8\right)a^3=280a^3\end{array}$

6- جد الحد الذي يحوي على x⁴ في مفكوك ${(1+x^2)}^6$ ثم جد معامله.

نفرض الحد الذي يحوي هوي على x⁴ هو r

$\begin{array}{rl}& Pr=C_{r-1}^n{x}^{n-r+1}{y}^{r-1}\\ & Pr=C_{r-1}^6(1{)}^{6-r+1}{\left(x^2\right)}^{r-1}\\ & Pr=C_{r-1}^6(1{)}^{7-r}{x}^{2r-2}\\ & x^{2r-2}=x^4⟹2r-2=4\\ & 2r=6]÷2⟹r=3\\ & P3=C_{3-1}^6(1{)}^{7-3}{x}^{6-2}=\frac{7\times 6}{2\times 1}{x}^4=15x^4\end{array}$

7- جد معامل x² في مفكوك ${(x^3+\frac{2}{x^2})}^9$

نفرض الحد الذي يحوي هوي على x² هو r

$\begin{array}{rl}& \mathrm{Pr}=C_{r-1}^n{x}^{n-r+1}{y}^{r-1}\\ & \mathrm{Pr}=C_{r-1}^9{\left(x^3\right)}^{9-r+1}{\left(\frac{2}{x^2}\right)}^{r-1}\\ & \mathrm{Pr}=C_{r-1}^9{\left(x^3\right)}^{10-r}(2{)}^{r-1}{\left(x^{-2}\right)}^{r-1}\\ & Pr=C_{r-1}^9{x}^{30-3r}(2{)}^{r-1}{x}^{-2r+2}\\ & x^{30-3r}\cdot x^{-2+2r}=x^2=x^{32-5r}=x^2\\ & 32-5r=2⟹5r=30]÷5⟹r=6\\ & P6=C_{6-1}^9{x}^{30-18}(2{)}^{6-1}{x}^{-12+2}\\ & P6=\frac{9\times 8\times 7\times 6\times 5}{5\times 4\times 3\times 2\times 1}{x}^{12}(2{)}^5{x}^{-10}=3\times 2\times 7\times 3\times 32x^2=4032x^2\end{array}$

8- جد الحد الخالي من x في مفكوك ${(x^2+\frac{2}{x^3})}^{10}$

نفرض الحد الخالي من x هو الذي يحوي على x⁰

^{$\begin{array}{rl}& \mathrm{Pr}=C_{r-1}^n{x}^{n-r+1}{y}^{r-1}\\ & \mathrm{Pr}=C_{r-1}^{10}{\left(x^2\right)}^{10-r+1}{\left(\frac{2}{x^3}\right)}^{r-1}=C_{r-1}^{10}{\left(x^2\right)}^{11-r}(2{)}^{r-1}{\left(x^{-3}\right)}^{r-1}\\ & Pr=C_{r-1}^{10}{x}^{22-2r}(2{)}^{r-1}{x}^{-3r+3}\\ & x^{22-2r}.x^{-3r+3}=x^0⟹x^{25-5r}=x^0\\ & 25-5r=0\Rightarrow 5r=5]÷5\Rightarrow r=5\\ & P5=C_{5-1}^{10}{x}^{22-10}(2{)}^{5-1}{x}^{-15+3}=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}{x}^{12}(16)x^{-12}=210(16)=3360\end{array}$}

9- جد قيمة $(99{)}^4$ باستخدام مبرهنة ذات الحدين.

$\begin{array}{rl}(99{)}^4& =(100-1{)}^4\\ & =C_0^4(100{)}^4-C_1^4(100{)}^3\left(1\right)+C_2^4\left(100{)}^2\right(1{)}^2-C_3^4\left(100\right)(1{)}^3+C_4^4(1{)}^4\\ & =100000000-4\left(1000000\right)+\frac{4\times 3}{2\times 1}\left(10000\right)-4\left(100\right)+1\\ & =100000000-4000000+60000-400+1=96059601\end{array}$

10- جد قيمة $(102{)}^4-(98{)}^4$

نأخذ ضعف الحدود الزوجية لأن الإشارة سالبة بين القوسين

$\begin{array}{rl}& (100+2{)}^4-(100-2{)}^4\\ & (100+2{)}^4=C_0^4(100{)}^4+C_1^4\left(100{)}^3\right(2)+C_2^4(100{)}^2(2{)}^2+C_3^4(100\left)\right(2{)}^3{C}_4^4(2{)}^4\\ & 2[P2+P4]\\ & =2\left[C_1^4\right(100{)}^3\left(2\right){+}_3^4\left(100\right)\left(2{)}^3\right]\\ & =2\left[4\right(1000000\left)\right(2)+4(100\left)\right(8\left)\right]\\ & =2\left[8\right(1000000)+32(100\left)\right]=2[8000000+3200]\\ & =16000000+6400=16006400\end{array}$

11- جد قيمة $(2+\sqrt{3}{)}^7+(2-\sqrt{3}{)}^7$

نأخذ ضعف الحدود الفردية لأن الإشارة موجبة بين القوسين

$\begin{array}{rl}& (2+\sqrt{3}{)}^7+(2-\sqrt{3}{)}^7\\ & \begin{array}{rl}(2+\sqrt{3}{)}^7=& C_0^7(2{)}^7+C_1^7(2{)}^6\left(\sqrt{3}\right)+C_2^7\left(2{)}^5\right(\sqrt{3}{)}^2+C_3^7\left(2{)}^4\right(\sqrt{3}{)}^3+C_4^7\left(2{)}^3\right(\sqrt{3}{)}^4\\ & +C_5^7\left(2{)}^2\right(\sqrt{3}{)}^5+C_6^7\left(2\right)(\sqrt{3}{)}^6+C_7^7(\sqrt{3}{)}^7\end{array}\\ & \begin{array}{c}2[P1+P3+P5+P7]\\ =2\left[C_0^7\right(2{)}^7+C_2^7\left(2{)}^5\right(\sqrt{3}{)}^2+C_4^7\left(2{)}^3\right(\sqrt{3}{)}^4+C_6^7\left(2\right)\left(\sqrt{3}{)}^6\right]\\ =\end{array}\\ & 2[128+\frac{7\times 6}{2\times 1}(32\left)\right(3)+\frac{7\times 6\times 5\times 4}{4\times 3\times 2\times 1}(8\left)\right(9)+7(2\left)\right(27\left)\right]\\ & =2[128+21(32\left)\right(3)+35(8\left)\right(9)+14(27\left)\right]\\ & =2[128+2016+2520+378]=2\left[5042\right]=10084\end{array}$