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2. Chứng minh rằng 12 +22 + 32 +......+n2 = \(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
HELPPPP 
\(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)(*)
Với n=1, ta có (*) luôn đúng
Giả sử (*) đúng với n=k ta có:
\(1^2+2^2+...+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\) (1)
Ta sẽ chứng minh (*) đúng với n=k+1, thật vậy từ (1) suy ra:
\(1^2+2^2+...+k^2+\left(k+1\right)^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)
\(=\left(k+1\right)\left[\frac{k\left(2k+1\right)}{6}+\left(k+1\right)\right]\)\(=\frac{\left(k+1\right)\left(2k^2+k+6k+6\right)}{6}\)
\(=\frac{\left(k+1\right)\left(2k^2+7k+6\right)}{6}=\frac{\left(k+1\right)\left(2k^2+4k+3k+6\right)}{6}\)
\(=\frac{\left(k+1\right)\left[2k\left(k+2\right)+3\left(k+2\right)\right]}{6}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)
Theo nguyên lý qui nạp (*) đúng với mọi n thuộc N*
Vậy ta có điều phải chứng minh
a) Ta có : \(f\left(x\right)+3f\left(\frac{1}{3}\right)=x^2\left(1\right)\Rightarrow f\left(\frac{1}{3}\right)+3f\left(\frac{1}{3}\right)=\left(\frac{1}{3}\right)^2\Leftrightarrow4f\left(\frac{1}{3}\right)=\frac{1}{9}\Leftrightarrow f\left(\frac{1}{3}\right)=\frac{1}{36}\)
Thay f(\(\frac{1}{3}\)) = \(\frac{1}{36}\) vào (1) được : \(f\left(x\right)=x^2-3f\left(\frac{1}{3}\right)=x^2-\frac{1}{12}\)
Vậy \(f\left(x\right)=x^2-\frac{1}{12}\)
b) \(f\left(x\right)+2f\left(\frac{1}{x}\right)=2x+\frac{1}{x}\) (2) . Thay \(x=\frac{1}{x}\) vào \(f\left(x\right)\) và \(f\left(\frac{1}{x}\right)\) được :
\(f\left(\frac{1}{x}\right)+2f\left(x\right)=\frac{2}{x}+x\) \(\Leftrightarrow2f\left(\frac{1}{x}\right)+4f\left(x\right)=\frac{4}{x}+2x\) (3)
Lấy (3) trừ (2) theo vế được: \(\left[2f\left(\frac{1}{x}\right)+4f\left(x\right)\right]-\left[f\left(x\right)+2f\left(\frac{1}{x}\right)\right]=\left(2x+\frac{4}{x}\right)-\left(2x+\frac{1}{x}\right)\)
\(\Leftrightarrow3f\left(x\right)=\frac{3}{x}\Leftrightarrow f\left(x\right)=\frac{1}{x}\)
c) \(f\left(x\right)+2f\left(-x\right)=x+1\) (4) . Thay x = -x vào f(x) và f(-x) được :
\(f\left(-x\right)+2f\left(x\right)=-x+1\Leftrightarrow2f\left(-x\right)+4f\left(x\right)=-2x+2\) (5)
Lấy (5) trừ (4) theo vế được :
\(\left[2f\left(-x\right)+4f\left(x\right)\right]-\left[f\left(x\right)+2f\left(-x\right)\right]=\left(-2x+2\right)-\left(x+1\right)\)
\(\Leftrightarrow3f\left(x\right)=-3x+1\Rightarrow f\left(x\right)=\frac{-3x+1}{3}\)
\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x^2-3x-2x+6}+\frac{1}{x^2-3x-4x+12}+\frac{1}{x^2-4x-5x+20}+\frac{1}{x^2-5x-6x+30}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{1-5}-\frac{1}{1-4}+\frac{1}{1-4}-\frac{1}{1-3}+\frac{1}{1-3}-\frac{1}{1-2}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{x^2-8x+12}=\frac{1}{8}\)
\(\Leftrightarrow x^2-8x+12=32\)
\(\Leftrightarrow\left(x-4\right)^2=36\)
\(\Leftrightarrow x=10\) hoặc \(x=-2\)
\(\frac{1}{x^2-5x+6}+\frac{1}{x^2-7x+12}+\frac{1}{x^2-9x+20}+\frac{1}{x^2-11x+30}=\frac{1}{8}\)\(\frac{1}{x^2-2x-3x+6}+\frac{1}{x^2-4x-3x+12}+\frac{1}{x^2-4x-5x+20}+\frac{1}{x^2-6x-5x+30}=\frac{1}{8}\)
\(\frac{1}{x\left(x-2\right)-3\left(x-2\right)}+\frac{1}{x\left(x-4\right)-3\left(x-4\right)}+\frac{1}{x\left(x-4\right)-5\left(x-4\right)}+\frac{1}{x\left(x-6\right)-5\left(x-6\right)}=\frac{1}{8}\)
\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)dhjjhhjhhjj
\(\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{8}\)
Còn lại tự giải quyết nha