Do \(x_1< x_2\). Do đó: \(x_1=\frac{2n-1-1}{2}=n-1\) và \(x_2=\frac{2n-1+1}{2}=n\)

Ta có \(x_1^2-2x_2+3=\left(n-1\right)^2-2n+3\)

\(=n^2-2n+1-2n+3=n^2-4n+4=\left(n-2\right)^2\ge0\)

Dấu "=" xảy ra <=> n=2

Vì x₁ < x₂.Do đó x₁=\(\frac{2n-1-1}{2}=n-1\)và x₂=\(\frac{2n-1+1}{2}=n\)

Ta có:\(x_{1_{ }}^{2^{ }^{ }}-2x_{2_{ }}+3=\left(n-1\right)^2-2n+3\)

\(=n^2-2n+1-2n+3=n^2-4n+4=\left(n-2\right)^2\ge0\)

Một người đều chơi 9 trận với 9 người người khác không có trận hòa. 

Do đó: \(x_1+y_1=x_2+y_2=....=x_{10}+y_{10}=9\)

Mà tổng số trận thắng bằng tổng số trận thua do đó:

\(x_1+x_2+...+x_{10}=y_1+y_2+y_3+...+y_{10}\)

Ta có: \(\left(x_1^2+x_2^2+...+x_{10}^2\right)-\left(y_1^2+y_2^2+...+y_{10}^2\right)\)

\(=\left(x_1^2-y_1^2\right)+\left(x_2^2-y_2^2\right)+.....+\left(x_{10}^2-y_{10}^2\right)\)

\(=9\left(x_1-y_1\right)+9\left(x_2-y_2\right)+....+9\left(x_{10}-y_{10}\right)\)

\(=9\left(x_1-y_1+x_2-y_2+....+x_{10}-y_{10}\right)\)

\(=9\left[\left(x_1+x_2+...+x_{10}\right)-\left(y_1+y_2+y_3+....+y_{10}\right)\right]=0\)

Vậy \(x_1^2+x_2^2+...+x_{10}^2=y_1^2+y_2^2+....+y_{10}^2\)

\(\frac{120}{x}+\frac{120}{x+10}=7\)

\(\Leftrightarrow\frac{120\left(x+10\right)}{x\left(x+10\right)}+\frac{120x}{x\left(x+10\right)}=\frac{7x\left(x+10\right)}{x\left(x+10\right)}\)

\(\Leftrightarrow120\left(x+10\right)+120x=7x\left(x+10\right)\)

\(\Leftrightarrow120x+1200+120x=7x^2+70x\)

\(\Leftrightarrow240x+1200=7x^2+70x\)

\(\Leftrightarrow240x+1200-7x^2-70x=0\)

\(\Leftrightarrow170x+1200-7x^2=0\)

\(\Leftrightarrow-7x^2+170x+1200=0\)

\(\Leftrightarrow7x^2+40x-210x-1200=0\)

\(\Leftrightarrow x\left(7x+40\right)-30\left(7x+40\right)=0\)

\(\Leftrightarrow\left(x-30\right)\left(7x+40\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=30\\7x=-40\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30\\x=-\frac{40}{7}\end{cases}}}\)

Áp dụng hệ thức Vi-ét,ta có : \(\hept{\begin{cases}x_1+x_2=-2\left(m-1\right)\\x_1x_2=4m\end{cases}}\)

Ta có : \(4x_1^2\left(1+x_2\right)+4x_2\left(1+x_1\right)+x_1^2x_2^2=36\)

\(\Rightarrow4\left(x_1^2+x_2^2\right)+4x_1x_2\left(x_1+x_2\right)+x_1^2x_2^2=36\)

\(\Rightarrow4\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+4x_1x_2\left(x_1+x_2\right)+x_1^2x_2^2=36\)

thay vào rồi tìm m thôi 

Ta có: \(x^2-5x+3=0\)

Áp dụng định lí viet ta có: \(\hept{\begin{cases}x_1+x_2=5\\x_1x_2=3\end{cases}}\)

a) \(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=5^2-2.3=19\)

b) \(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3\left(x_1+x_2\right)x_1x_2=5^3-3.5.3=80\)

c) \(C=\left|x_1-x_2\right|\)>0

=> \(C^2=x_1^2+x_2^2-2x_1x_2=19-2.3=13\)

=> C = căn 13

d) \(D=x_2+\frac{1}{x_1}+x_1+\frac{1}{x_2}=\left(x_1+x_2\right)+\frac{x_1+x_2}{x_1x_2}=5+\frac{5}{3}=5\frac{5}{3}\)

e) \(E=\frac{1}{x_1+3}+\frac{1}{x_2+3}=\frac{\left(x_1+x_2\right)+6}{x_1x_2+3\left(x_1+x_2\right)+9}=\frac{5+6}{3+3.5+9}=\frac{11}{27}\)

g) \(G=\frac{x_1-3}{x_1^2}+\frac{x_2-3}{x_2^2}=\left(\frac{1}{x_1}+\frac{1}{x_2}\right)-3\left(\frac{1}{x_1^2}+\frac{1}{x_2^2}\right)\)

\(=\frac{x_1+x_2}{x_1x_2}-3\frac{x_1^2+x_2^2}{x_1^2.x_2^2}=\frac{5}{3}-3.\frac{19}{3^2}=-\frac{14}{3}\)