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A=1/2^2 + 1/3^2 + 1/4^2 + ... + 1/2017^2
A < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2016.2017
A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2016 - 1/2017
A < 1 - 1/2017 < 1 (1)
B = 2!/3! + 2!/4! + 2!/5! + ... + 2!/2017!
B = 2!.(1/3! + 1/4! + 1/5! + ... + 1/2017!)
B < 2.(1/2.3 + 1/3.4 + 1/4.5 + ... + 1/2016.2017)
B < 2.(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/2016 - 1/2017)
B < 2.(1/2 - 1/2017) < 2.1/2 = 1 (2)
Từ (1) và (2) => A + B < 2 (đpcm)
Câu 1:
\(F=\frac{\frac{x^3-x}{x+1}+\frac{2x-2}{1+\frac{x}{2}}}{\frac{x^3-3x^2}{x-3}-\frac{2x^2+8}{x+2}}\left(ĐKXĐ:x\ne3;-2;-1\right)\)
\(F=\frac{\frac{x\left(x-1\right)\left(x+1\right)}{x+1}+\frac{2x-2}{1+\frac{x}{2}}}{\frac{x^2\left(x-3\right)}{x-3}-\frac{2x^2+8}{x+2}}\)
\(F=\frac{\frac{\left(x^2-x\right)\left(1+\frac{x}{2}\right)+2x-2}{1+\frac{x}{2}}}{\frac{x^2\left(x+2\right)-2x^2-8}{x+2}}\)
\(F=\frac{\frac{x^2+\frac{x^3}{2}-x-\frac{x^2}{2}+2x-2}{1+\frac{x}{2}}}{\frac{x^3-8}{x+2}}\)
\(F=\frac{\frac{x^2}{2}+\frac{x^3}{2}+x-2}{1+\frac{x}{2}}.\frac{x+2}{x^3-8}\)
Câu 2:
\(G=\frac{\frac{x^4+1}{x^3-1}-x}{\frac{x}{x^2+x+1}-\frac{2}{x-1}}\left(ĐKXĐ:x\ne1\right)\)
\(G=\frac{\frac{x^4+1-x\left(x^3-1\right)}{x^3-1}}{\frac{x\left(x-1\right)-2\left(x^2+x+1\right)}{x^3-1}}\)
\(G=\frac{x+1}{x^3-1}:\frac{x^2-x-2x^2-2x-2}{x^3+1}\)
\(G=\frac{x+1}{-x^2-3x-2}\)
\(G=\frac{x+1}{-\left(x+2\right)\left(x+1\right)}\)
\(G=-\frac{1}{x+2}\)Tại x=2017 ta đc:\(G=-\frac{1}{2+2017}=-\frac{1}{2019}\)
a: \(=\dfrac{4}{x+2}-\dfrac{3}{x-2}+\dfrac{12}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8-3x-6+12}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
b: \(=\dfrac{6x+3\left(x-1\right)+2\left(x-2\right)}{6}=\dfrac{6x+3x-3+2x-4}{6}=\dfrac{11x-7}{6}\)
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
@Vũ Minh Tuấn