Giải bài 1 trang 57 – SGK môn Đại số và Giải tích lớp 11

Hướng dẫn:

Công thức Nhị thức Niu - tơn:

$(a+b)^n=C^0_na^n+C^1_na^{n-1}b+C^2_na^{n-2}b^2+...+C^n_nb^n$

Theo công thức nhị thức Niu-tơn ta có:

a) 

$(a+2b)^5=C_5^0a^5+C_5^1a^4(2b)^1+C_5^2a^3(2b)^2+C_5^3a^2(2b)^3+C_5^4a^1(2b)^4+C_5^5(2b)^5\\ =a^5+10a^4b+40a^3b^2+80a^2b^3+80ab^4+32b^5$

b) 

$(a-\sqrt{2})^6=C_6^0a^6+C_6^1a^5(-\sqrt{2})^1+C_6^2a^4(-\sqrt{2})^2+C_6^3a^3(-\sqrt{2})^3+C_6^4a^2(-\sqrt{2})^4+C_6^5a^1(-\sqrt{2})^5+C_6^6(-\sqrt{2})^6 \\ =a^6-6a^5\sqrt{2}+15a^4(\sqrt{2})^2-20a^3(\sqrt{2})^3+15a^2(\sqrt{2})^4-6a(\sqrt{2})^5+(\sqrt{2})^6\\ =a^6-6\sqrt{2}a^5+30a^4-40\sqrt{2}a^3+60a^2-24\sqrt{2}a+8$

c)

$\begin{aligned} & {{\left( x-\frac{1}{x} \right)}^{13}}=C_{13}^{0}{{x}^{13}}+C_{13}^{1}{{x}^{12}}{{\left( \frac{-1}{x} \right)}^{1}}+C_{13}^{2}{{x}^{11}}{{\left( \frac{-1}{x} \right)}^{2}}+...+C_{13}^{12}{{x}^{1}}{{\left( \frac{-1}{x} \right)}^{12}}+C_{13}^{13}{{\left( \frac{-1}{x} \right)}^{13}} \\ & =C_{13}^{0}{{x}^{13}}+C_{13}^{1}{{x}^{12}}{{\left( -{{x}^{-1}} \right)}^{1}}+C_{13}^{2}{{x}^{11}}{{\left( -{{x}^{-1}} \right)}^{2}}+...+C_{13}^{12}{{x}^{1}}{{\left( -{{x}^{-1}} \right)}^{12}}+C_{13}^{13}+{{\left( {{-x}^{-1}} \right)}^{13}} \\ & =\sum\limits_{k=0}^{13}{C_{13}^{k}{{x}^{13-k}}{{\left( -x \right)}^{-k}}=}\sum\limits_{k=0}^{13}{C_{13}^{k}{{\left( -1 \right)}^{k}}{{x}^{13-2k}}} \\ \end{aligned}$