Viết các khai triển sau : a, P(x)=(3x-1)^4 b,p(x)= (1-2x)^5 c,p(x)=(x+1/x)^6 21/09/2021 Bởi Allison Viết các khai triển sau : a, P(x)=(3x-1)^4 b,p(x)= (1-2x)^5 c,p(x)=(x+1/x)^6
Đáp án: Giải thích các bước giải: \[\begin{array}{l} a.P = {(3x – 1)^4} = \sum\limits_{k = 0}^4 {C_4^k{{(3x)}^{4 – k}}.{{( – 1)}^k}} = 81{x^4} – 108{x^3} + 54{x^2} – 12x + 1\\ b.P = {(1 – 2x)^5} = \sum\limits_{k = 0}^5 {C_5^k{{(1)}^{5 – k}}.{{( – 2x)}^k}} = 1 – 10x + 40{x^2} – 80{x^3} + 80{x^4} – 32{x^5}\\ c.P = {(x + \frac{1}{x})^6} = {(x + {x^{ – 1}})^6} = \sum\limits_{k = 0}^6 {C_6^k} .{x^{6 – k}}.{({x^{ – 1}})^k} = \sum\limits_{k = 0}^6 {C_6^k} .{x^{6 – 2k}} = {x^6} + 6{x^4} + 15{x^2} + 20 + 15{x^{ – 2}} + 6{x^{ – 4}} + {x^{ – 6}} = {x^6} + 6{x^4} + 15{x^2} + 20 + \frac{{15}}{{{x^2}}} + \frac{6}{{{x^4}}} + \frac{1}{{{x^6}}} \end{array}\] Bình luận
Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
a.P = {(3x – 1)^4} = \sum\limits_{k = 0}^4 {C_4^k{{(3x)}^{4 – k}}.{{( – 1)}^k}} = 81{x^4} – 108{x^3} + 54{x^2} – 12x + 1\\
b.P = {(1 – 2x)^5} = \sum\limits_{k = 0}^5 {C_5^k{{(1)}^{5 – k}}.{{( – 2x)}^k}} = 1 – 10x + 40{x^2} – 80{x^3} + 80{x^4} – 32{x^5}\\
c.P = {(x + \frac{1}{x})^6} = {(x + {x^{ – 1}})^6} = \sum\limits_{k = 0}^6 {C_6^k} .{x^{6 – k}}.{({x^{ – 1}})^k} = \sum\limits_{k = 0}^6 {C_6^k} .{x^{6 – 2k}} = {x^6} + 6{x^4} + 15{x^2} + 20 + 15{x^{ – 2}} + 6{x^{ – 4}} + {x^{ – 6}} = {x^6} + 6{x^4} + 15{x^2} + 20 + \frac{{15}}{{{x^2}}} + \frac{6}{{{x^4}}} + \frac{1}{{{x^6}}}
\end{array}\]