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30 tháng 12 2018

\(\frac{x^{24}+x^{20}+...+x^4+1}{x^{26}+x^{24}+...+x^2+1}=\frac{x^{24}+x^{20}+...+x^4+1}{\left(x^{24}+x^{20}+...+x^4+1\right)+\left(x^{26}+x^{22}+...+x^2\right)}\)

\(=1-\frac{x^2\left(x^{24}+x^{20}+...+x^4+x^1\right)}{\left(1+x^2\right)\left(x^{24}+2^{20}+...+x^4+1\right)}=1-\frac{x^2}{1+x^2}\)

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

30 tháng 12 2018

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: \(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)

\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)

\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)

\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{24}+x^{20}+x^{16}+...+1\right)\left(x^2+1\right)}\)

\(=\dfrac{1}{x^2+1}\)

22 tháng 2 2021

 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)

27 tháng 10 2018

\(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)

\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+x^{18}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)

\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+x^{16}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)

\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)

\(=\dfrac{1}{x^2+1}\)

18 tháng 1 2017

\(A=\frac{x^{24}+x^{20}+x^{16}+....+x^4+1}{x^{26}+x^{24}+x^{22}+.....+x^2+1}\) (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=\(\frac{1}{x^2+1}\)

10 tháng 1 2018

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!

10 tháng 1 2018

khiếp, dài tek

Cái đề rõ ngắn giải thì rõ dài

~ Nể sự kiên nhẫn của bà thiệt = = ~

1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)

\(=x^2+4x+4+x^2-6x+9\)

\(=2x^2-2x+13\)

2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)

\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)

\(=-2x+7\)

3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)

\(=x^2-25-x^2-10x-25\)

=-10x-50

4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)

\(=x^2-6x+9-x^2+16\)

=-6x+25

5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)

\(=y^2-6y+9-y^2+6y-9\)

=0

6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)

\(=4x^2+12x+9-4x^2+9\)

=12x+18

a: \(=\dfrac{2^{19}\cdot3^9+2^{20}\cdot3^{10}}{2^{19}\cdot3^9+2^{18}\cdot3^9\cdot5}=\dfrac{2^{19}\cdot3^9\left(1+2\cdot3\right)}{2^{18}\cdot3^9\left(2+5\right)}=2\)