Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{24}+x^{20}+x^{16}+...+1\right)\left(x^2+1\right)}\)
\(=\dfrac{1}{x^2+1}\)
x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1
=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1(x24+x20+x16+...+1)(x2+1)
a: \(=\dfrac{2^{19}\cdot3^9+2^{20}\cdot3^{10}}{2^{19}\cdot3^9+2^{18}\cdot3^9\cdot5}=\dfrac{2^{19}\cdot3^9\left(1+2\cdot3\right)}{2^{18}\cdot3^9\left(2+5\right)}=2\)
\(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+x^{18}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+x^{16}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)
\(=\dfrac{1}{x^2+1}\)
Rút gọn phân thức:
A=\(\frac{x^{24}+x^{20}+x^{16}+.....+x^4+1}{x^{26}+x^{24}+x^{22}+.......+x^2+1}\)
\(A=\frac{x^{24}+x^{20}+x^{16}+....+x^4+1}{x^{26}+x^{24}+x^{22}+.....+x^2+1}\) (1)
Ta có \(x^{26}+x^{24}+x^{22}+...+x^2+1\)
\(=\left(x^{26}+x^{22}+x^{18}+....+x^2\right)+\left(x^{24}+x^{20}+...+x^4+1\right)\)
\(=x^2\left(x^{24}+x^{20}+.....+x^4+1\right)+\left(x^{24}+x^{20}+...+x^4+1\right)\)
\(=\left(x^2+1\right)\left(x^{24}+x^{20}+x^{16}+....+x^4+1\right)\) (2)
Từ (1),(2) ta có \(A=\frac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+x^{16}+....+x^4+1\right)}=\frac{1}{x^2+1}\)
Vậy A=\(\frac{1}{x^2+1}\)
\(\frac{x^{24}+x^{20}+...+x^4+1}{x^{26}+x^{24}+...+x^2+1}=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^{24}+x^{20}+...+x^4+1\right)+\left(x^{26}+x^{22}+...+x^2\right)}\)
\(=1-\frac{x^2\left(x^{24}+x^{20}+...+x^4+x^1\right)}{\left(1+x^2\right)\left(x^{24}+2^{20}+...+x^4+1\right)}=1-\frac{x^2}{1+x^2}\)
\(=\frac{1+x^2-x^2}{1+x^2}=\frac{1}{1+x^2}\)
Hoặc cách khác:
\(\frac{x^{24}+x^{20}+...+x^4+1}{x^{26}+x^{24}+...+x^2+1}=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^{24}+x^{20}+...+x^4+1\right)+x^2\left(x^4+x^{20}+...+x^4+1\right)}\)
\(=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+...+x^4+1\right)}=\frac{1}{x^2+1}\)
Ta có:
\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)
Xét \(M=x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1\)
\(\Rightarrow x^4M=x^{28}+x^{24}+x^{20}+x^{16}+...+x^8+x^4\)
\(\Rightarrow x^4M-M=\left(x^{28}+x^{24}+x^{20}+...+x^8+x^4\right)-\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)\)
\(\Rightarrow\left(x^4-1\right)M=x^{28}-1\)
\(\Rightarrow M=\dfrac{x^{28}-1}{x^4-1}\)
Xét \(N=x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1\)
\(\Rightarrow x^2N=x^{28}+x^{26}+x^{24}+x^{20}+...+x^4+x^2\)
\(\Rightarrow x^2N-N=\left(x^{28}+x^{26}+x^{24}+...+x^4+x^2\right)-\left(x^{26}+x^{24}+x^{22}+...+x^2+1_{ }\right)\)
\(\Rightarrow\left(x^2-1\right)N=x^{28}-1\)
\(\Rightarrow N=\dfrac{x^{28}-1}{x^2-1}\)
Ta có:
\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)
\(=\dfrac{M}{N}=\dfrac{\dfrac{x^{28}-1}{x^4-1}}{\dfrac{x^{28}-1}{x^2-1}}\)
\(=\dfrac{x^{28}-1}{x^4-1}.\dfrac{x^2-1}{x^{28}-1}=\dfrac{x^2-1}{x^4-1}\)
\(=\dfrac{x^2-1}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{1}{x^2+1}\)
Chúc bạn học tốt!
khiếp, dài tek
Cái đề rõ ngắn giải thì rõ dài
~ Nể sự kiên nhẫn của bà thiệt = = ~