Tham khảo:

\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{121}-1\right)\)

\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{121}\right)\)

\(-A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{120}{121}\)

\(-A=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot10\cdot12}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot11\cdot11}\)

\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot10\right)\left(3\cdot4\cdot5\cdot...\cdot12\right)}{\left(2\cdot3\cdot4\cdot...\cdot11\right)\left(2\cdot3\cdot4\cdot...\cdot11\right)}\)

\(-A=\frac{1\cdot12}{11\cdot2}=\frac{6}{11}\)

\(A=-\frac{6}{11}\)

\(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{37}-\frac{1}{38}\)

\(B=1-\frac{1}{38}=\frac{37}{38}\)

Đặt \(g(x)=10x\).

Ta có \(g\left(1\right)=10=f\left(1\right);g\left(2\right)=20=f\left(2\right);g\left(3\right)=30=f\left(3\right)\).

Từ đó \(\left\{{}\begin{matrix}f\left(1\right)-g\left(1\right)=0\\f\left(2\right)-g\left(2\right)=0\\f\left(3\right)-g\left(3\right)=0\end{matrix}\right.\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=Q\left(x\right).\left(x-1\right)\left(x-2\right)\left(x-3\right)\).

\(\Rightarrow f\left(x\right)=10x+Q\left(x\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

\(\Rightarrow f\left(8\right)+f\left(-4\right)=80+Q\left(x\right).7.6.5+\left(-40\right)+Q\left(x\right).\left(-5\right).\left(-6\right).\left(-7\right)=80-50=40\).

Ta có: a + 3c + a + 2b = 2019 + 2020 = 4039 

=> 2 ( a + b + c ) = 4039 - c (1)

a; b ; c là các số hữu tỉ không âm => a; b ; c \(\ge\)0 

=> 2 ( a + b + c ) = 4039 - c \(\le\)4039 

=> a + b + c \(\le\frac{4039}{2}=2019\frac{1}{2}\)

mà f(1) = a + b + c 

=> f (1) \(\le2019\frac{1}{2}\)

Dấu "=" xảy ra <=> c = 0 ; a = 2019 ; b = 1/2