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=>\(\frac{B}{2^2}\)=\(\frac{1}{2^2}\)\(\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
=> \(\frac{B}{4}=\frac{1}{4}.A\)
=>A=B
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(6-1\right)}\)
\(=\frac{2.6}{3.5}=\frac{4}{5}\)
\(\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=...=\frac{a9}{a1}\)
\(\Rightarrow\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=...=\frac{a9}{a1}=\frac{a1+a2+a3+...+a9}{a2+a3+a4+...+a1}=1\)
\(\Rightarrow a1=a2,a2=a3,a3=a4,...,a9=a1\)
\(\Rightarrow a1=a2=a3=a4=a5=...=a9\)
Mà a1=5\(\Rightarrow a5=5\)
\(A=1+2+2^2+2^3+...+2^{100}\)(có 101 số)
\(A=1+\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(A=1+2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)
\(A=1+2\cdot31+2^6\cdot31+...+2^{96}\cdot31\)
\(A=1+31\left(2+2^6+...+2^{96}\right)\)
\(\Rightarrow A:31\) dư 1
f(3)=-2.3+7,3=5=5 ( không thỏa đề bài)
\(\Rightarrow f\left(3\right)\in\varnothing\)
f(3)=3+9,3=12,2\(\ge5\)( thỏa đề bài)
Vậy f(3)=12,2