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A = 25^28 + 25^24 + .... + 25^4 + 1
25^4.A = 25^32 + 25^28 + 25^24 + ...+ 25^4
=> (25^4 - 1)A = 25^32 - 1
=> A = (25^32 - 1)/(25^4 - 1)
B = 25^30 + 25^28 + ... + 25^2 + 1
25^2.B = 25^32 + 25^30 + 25^28 + ...+ 25^2
=> (25^2 - 1)B = 25^32 - 1
=> B = (25^32 - 1)/(25^2 - 1)
vậy :
A/B = [(25^32 - 1)/(25^4 - 1)]/[(25^32 - 1)/(25^2 - 1)] = (25^2 - 1)/(25^4 -1) = 1/(25^2 + 1) = 1/626
Địa chỉ chi tiết : yahoo.com
\(A=\frac{25^{28}+25^{24}+...+25^4+25^0}{25^{30}+25^{28}+...+25^2+25^0}\)
\(=\frac{25^{28}+25^{24}+...+25^0}{\left(25^{28}+25^{24}+...+25^0\right)+\left(25^{30}+23^{26}+...+25^2\right)}\)
\(=\frac{25^{28}+25^{24}+...+25^0}{\left(25^{28}+25^{24}+...+25^0\right)+25^2\left(25^{28}+23^{24}+...+25^0\right)}\)
\(=\frac{25^{28}+25^{24}+...+25^0}{\left(25^{28}+25^{24}+...+25^0\right)\left(1+25^2\right)}\)
\(=\frac{1}{1+25^2}\)
\(=\frac{1}{626}\)
Vì \(\left|x-y\right|\ge0;\left|y+\frac{9}{25}\right|\ge0\Rightarrow\left|x-y\right|+\left|y+\frac{9}{25}\right|\ge0\)
Mà \(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)
\(\Rightarrow\left|x-y\right|=0;\left|y+\frac{9}{25}\right|=0\)
\(\left|y+\frac{9}{25}\right|=0\Rightarrow y=\frac{-9}{25}\)
\(\Rightarrow\left|x-y\right|=\left|x-\frac{-9}{25}\right|=0\Rightarrow x=\frac{-9}{25}\)
C = \(\frac{25^{28}+25^{24}+...+25^4+25^0}{25^{30}+25^{28}+...+25^2+25^0}\)
= \(\frac{25^{28}+25^{24}+...+25^2+25^0}{\left(25^{30}+25^{26}+...+25^2\right)+\left(25^{28}+25^{24}+...+25^0\right)}\)
= \(\frac{25^{28}+25^{24}+...+25^0}{25^2\left(25^{28}+25^{24}+...+1\right)+\left(25^{28}+25^{24}+...+1\right)}\)
= \(\frac{25^{28}+25^{24}+...+1}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}=\frac{1}{626}\)
Cho 10 ****