B=\(\frac{5^{100}+7}{5^{100}+5}\)=\(\frac{5^{100}+6+1}{5^{100}+4+1}\)

Mà \(\frac{5^{100}+6+1}{5^{100}+4+1}\)>\(\frac{5^{100}+6}{5^{100}+4}\)

\(\Rightarrow\)B>A

Áp dụng a/b > 1 => a/b > a+m/b+m (a,b,m thuộc N*)

=> \(\frac{5^{100}+6}{5^{100}+4}>\frac{5^{100}+6+1}{5^{100}+4+1}\)

                       \(>\frac{5^{100}+7}{5^{100}+5}\)

A=5^100+6=5^100+5+1=5^101+1 ; 5^100+4=5^100+5-1=5^101-1
B=5^100+7=5^100+5+2=5^101+2 ; 5^100+5=5^101
vs nha
 

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`Answer:`

\(A=\left(1^{100}+2^{100}+...+10^{100}\right)\left(5^{10}-25^5\right)\)

\(=\left(1^{100}+2^{100}+...+10^{100}\right)[5^{10}-\left(5^2\right)^5]\)

\(=\left(1^{100}+2^{100}+...+10^{100}\right)\left(5^{10}-5^{10}\right)\)

\(=\left(1^{100}+2^{100}+...+10^{100}\right).0\)

\(=0\)

\(B=\left(2^5+3^5+4^5\right)\left(1^2+2^2+...+100^2\right)\left(4^{10}-2^{20}\right)\)

\(=\left(2^5+3^5+4^5\right)\left(1^2+2^2+...+100^2\right)\left(2^{2.10}-2^{20}\right)\)

\(=\left(2^5+3^5+4^5\right)\left(1^2+2^2+...+100^2\right).0\)

\(=0\)

\(A=\dfrac{100^{100}-1}{100^{100}-5}=\dfrac{\left(100^{100}-1\right)\left(100^{100}+1\right)}{\left(100^{100}-5\right)\left(100^{100}+1\right)}=\dfrac{100^{200}-1}{\left(100^{100}-5\right)\left(100^{100}+1\right)}\)

\(B=\dfrac{100^{100}+5}{100^{100}+1}=\dfrac{\left(100^{100}+5\right)\left(100^{100}-5\right)}{\left(100^{100}-5\right)\left(100^{100}+1\right)}=\dfrac{100^{200}-25}{\left(100^{100}-5\right)\left(100^{100}+1\right)}\)

\(\Rightarrow A>B\)

-1

Đó là :

5 x 5 x 5 - 5 x 5 = 100

mik đi cùng họ

dấu nhân nha bn

60123