\(\frac{149-x}{26}+\frac{171-x}{24}+\frac{189-x}{22}+\frac{203-x}{20}=10\)
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MT
23 tháng 6 2015
\(\frac{74-x}{26}+\frac{75-x}{25}+\frac{76-x}{24}+\frac{77-x}{23}+\frac{78-x}{22}=-5\)
\(\frac{74-x}{26}+1+\frac{75-x}{25}+1+\frac{76-x}{24}+1+\frac{77-x}{23}+1+\frac{78-x}{22}=-5+5\)
\(\frac{74-x}{26}+\frac{26}{26}+\frac{75-x}{25}+\frac{25}{25}+\frac{76-x}{24}+\frac{24}{24}+\frac{77-x}{23}+\frac{23}{23}+\frac{78-x}{22}+\frac{22}{22}=0\)
\(\frac{100-x}{26}+\frac{100-x}{25}+\frac{100-x}{24}+\frac{100-x}{23}+\frac{100-x}{22}=0\)
\(\left(100-x\right)\left(\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{23}+\frac{1}{22}\right)=0\)
=>100-x=0 ( \(\left(\frac{1}{26}+\frac{1}{25}+\frac{1}{24}+\frac{1}{23}+\frac{1}{22}\right)\ne0\))
x=100
hahaha
DK
13 tháng 4 2016
Ta nhận thấy mẫu của biểu thức trên là:
x26+x24+x22+...+x2+1=(x26+x22+...+x2)+(x24+x20+...+x4+1)
=x2(x24+x20+...+x16+...+1)+(x24+x20+...+x4+1)
=(x24+x20+...+1)(x2+1)
Như vậy\(\frac{x^{24}+x^{20}+x^{16}+...+1}{\left(x^{24}+x^{20}+...+1\right)\left(x^2+1\right)}\)=\(\frac{1}{x^2+1}\)
15 tháng 3 2017
\(B=\frac{1+x^2+x^4+...+x^{26}}{1+x^4+x^8+...+x^{24}}\)
\(=\frac{\frac{\left(x^2-1\right)\left(1+x^2+x^4+...+x^{26}\right)}{x^2-1}}{\frac{\left(x^4-1\right)\left(1+x^4+x^8+...+x^{24}\right)}{x^4-1}}\)
\(=\frac{\frac{x^{28}-1}{x^2-1}}{\frac{x^{28}-1}{x^4-1}}=\frac{x^4-1}{x^2-1}=x^2+1\)
AN
27 tháng 10 2018
\(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+x^{18}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+x^{16}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^2+1\right)\left(x^{24}+x^{20}+x^{16}+...+1\right)}\)
\(=\dfrac{1}{x^2+1}\)
ND
Rút gọn phân thức:
A=\(\frac{x^{24}+x^{20}+x^{16}+.....+x^4+1}{x^{26}+x^{24}+x^{22}+.......+x^2+1}\)
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=\(\frac{1}{x^2+1}\)
3 tháng 7 2017
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
\(\frac{149-x}{26}+\frac{171-x}{24}+\frac{189-x}{22}+\frac{203-x}{20}=10\)
\(\left(\frac{149-x}{26}-1\right)+\left(\frac{171-x}{24}-2\right)+\left(\frac{189-x}{22}-3\right)+\left(\frac{203-x}{20}-4\right)=0\)
\(\frac{123-x}{26}+\frac{123-x}{24}+\frac{123-x}{22}+\frac{123-x}{20}=0\)
\(\left(123-x\right)\left(\frac{1}{26}+\frac{1}{24}+\frac{1}{22}+\frac{1}{20}\right)=0\)
\(123-x=0\left(vì\frac{1}{26}+\frac{1}{24}+\frac{1}{22}+\frac{1}{20}\ne0\right)\)
\(\Rightarrow x=123\)