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Ta có : x1 + x2 + x3 + x4 +...... + x50 + x51 = 0
<=> (x1 + x2) + (x3 + x4) +...... + (x49 + x50) + x51
<=> 1 + 1 + 1 + ..... + 1 + x51 = 0
=> 50 + x51 = 0
=> x51 = -50
\(\frac{x_1-1}{2010}=...=\frac{x_{2010}-2010}{1}=\frac{x_1+x_2+...+x_{2010}-\left(1+2+...+2010\right)}{2010+2009+...+1}\)
\(=\frac{2\left(1+2+...+2010\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=1\)
Vậy thay vào ta được: \(x_1=x_2=...=x_{2010}=2011\)
\(\frac{x_1-1}{2010}=\frac{x_2-2}{2009}=...=\frac{x_{2010}-2010}{1}=\frac{\left(x_1-1\right)+\left(x_2-2\right)+...+\left(x_{2010}-2010\right)}{1+2+...+2010}\) (TC DTSBN)
\(=\frac{\left(x_1+x_2+...+x_{2010}\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=\frac{2.\left(1+2+...+2010\right)-\left(1+2+...+2010\right)}{1+2+...+2010}=1\)
\(\Rightarrow x_1-1=2010;x_2-1=2009;....;x_{2010}-2010=1\)
=> x1 = x2 = x3 =..... = x2010 = 2011
Ta có ; \(\frac{1-x_1}{99}=\frac{2-x_2}{98}=\frac{3-x_3}{97}=...=\frac{99-x_{99}}{1}\)
\(\Leftrightarrow\frac{1-x_1}{99}+1=\frac{2-x_2}{98}+1=\frac{3-x_3}{97}+1=...=\frac{99-x_{99}}{1}+1\)
\(\Leftrightarrow\frac{100-x_1}{99}=\frac{100-x_2}{98}=\frac{100-x_3}{97}=...=\frac{100-x_{99}}{1}\)
Áp dụng t/c dãy tỉ số bằng nhau : \(\frac{100-x_1}{99}=\frac{100-x_2}{98}=\frac{100-x_3}{97}=...=\frac{100-x_{99}}{1}\)
\(=\frac{\left(100-x_1\right)+\left(100-x_2\right)+\left(100-x_3\right)=...=\left(100-x_{99}\right)}{1+2+3+...+98+99}\)
\(=\frac{100.99-\left(x_1+x_2+x_3+...+x_{99}\right)}{1+2+3+...+99}=\frac{100.99-4950}{\frac{99.100}{2}}=1\)
\(\Rightarrow x_i=100-\left(100-i\right)=i\)với \(i=1,2,3,...,99\)
\(\frac{1-x_1}{99}=\frac{2-x_2}{98}=\frac{3-x_3}{97}=...=\frac{99-x_{99}}{1}=\)\(\frac{\left(1+2+3+..+99\right)-\left(x_1+x_2+x_3+...+x_{99}\right)}{99+98+97+...+1}\)\(=\frac{4950-4950}{4950}=0\)
\(\Rightarrow1-x_1=2-x_2=3-x_3=...=99-x_{99}=0\)
\(\Rightarrow x_i=i-0\left(i=1,2,3,...,99\right)\)