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Theo Vi-ét cho 3 số (chứng minh bằng hệ số bất định)
\(\hept{\begin{cases}x_1+x_2+x_3=0\\x_1x_2+x_2x_3+x_1x_3=-3\\x_1x_2x_3=-1\end{cases}}\)
\(A=\frac{1+2x_1}{1+x_1}+\frac{1+2x_2}{1+x_2}+\frac{1+2x_3}{1+x_3}\)
\(=3+\frac{x_1}{1+x_1}+\frac{x_2}{1+x_2}+\frac{x_3}{1+x_3}\)
\(=3+\frac{x_1\left(1+x_2\right)\left(1+x_3\right)+x_2\left(1+x_1\right)\left(1+x_3\right)+x_3\left(1+x_1\right)\left(1+x_2\right)}{\left(1+x_1\right)\left(1+x_2\right)\left(1+x_3\right)}\)
\(=3+\frac{x_1\left(1+x_2+x_3+x_2x_3\right)+x_2\left(1+x_1+x_3+x_1x_3\right)+x_3\left(1+x_1+x_2+x_1x_2\right)}{\left(1+x_1+x_2+x_1x_2\right)\left(1+x_3\right)}\)
\(=3+\frac{\left(x_1+x_2+x_3\right)+2\left(x_1x_2+x_2x_3+x_3x_1\right)+3x_1x_2x_3}{1+x_1+x_2+x_3+x_1x_2+x_1x_3+x_2x_3+x_1.x_2.x_3}\)
\(=3+\frac{0+2.\left(-3\right)+3.\left(-1\right)}{1+0-3-1}\)
\(=6\)
Do x1 là một nghiệm của đa thức f(x) nên ta có: \(x_1^3-3x_1+1=0\)
\(\Leftrightarrow\)\(\left(x_1+1\right)\left(x_1^2-x_1+1\right)=3x_1\)\(\Leftrightarrow\)\(x_1+1=\frac{3x_1}{x_1^2-x_1+1}\)
Có: \(A==\frac{1+2x_1}{1+x_1}+\frac{1+2x_2}{1+x_2}+\frac{1+2x_3}{1+x_3}=3+\left(\frac{x_1}{1+x_1}+\frac{x_2}{1+x_2}+\frac{x_3}{1+x_3}\right)\)
\(A=3+\left(\frac{x_1\left(x_1^2-x_1+1\right)}{3x_1}+\frac{x_2\left(x^2_2-x_2+1\right)}{3x_2}+\frac{x_3\left(x_3^2-x_3+1\right)}{3x_3}\right)\)
\(A=3+\frac{\left(x_1^2+x_2^2+x_3^2\right)-\left(x_1+x_2+x_3\right)+3}{3}\)
\(A=3+\frac{\left(x_1+x_2+x_3\right)^2-2\left(x_1x_2+x_2x_3+x_3x_1\right)-\left(x_1+x_2+x_3\right)+3}{3}\)
Đến đây theo Vi-et bậc 3
\(\hept{\begin{cases}x_1+x_2+x_3=0\\x_1x_2+x_2x_3+x_3x_1=-3\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Với n=2
=> \(x_1+\frac{1}{x_1}=x_2+\frac{1}{x_2}\)
\(\Rightarrow x_1-x_2=\frac{1}{x_1}-\frac{1}{x_2}\)
\(\Rightarrow\left(x_1-x_2\right)-\frac{x_1-x_2}{x_1x_2}=0\)
\(\Rightarrow\left(x_1-x_2\right)\left(1-\frac{1}{x_1x_2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x_1-x_2=0\\1-\frac{1}{x_1x_2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x_1=x_2\\x_1x_2=1\end{cases}}}\)
*) n=k
=> \(x_1+\frac{1}{x_1}=x_2+\frac{1}{x_2}=...=x_k+\frac{1}{x_k}\)
thì \(x_1=x_2=x_3=...=x_k\)hoặc \(\left|x_1x_2...x_k\right|=0\)
Với n=k+1
=> \(x_1+\frac{1}{x_1}=x_2+\frac{1}{x_2}=x_3+\frac{1}{x_3}=...x_{k+1}+\frac{1}{x_1}\)
=> \(x_1+\frac{1}{x_2}=x_2+\frac{1}{x_3}=....=x_k+\frac{1}{x_{k+1}}=x_{k+1}+\frac{1}{x_1}\)
\(\Rightarrow x_{k-1}+\frac{1}{x_k}=x_k+\frac{1}{x_1}=x_{k+1}+\frac{1}{x_1}\)
\(\Rightarrow x_k-x_{k+1}=0\)
\(\Rightarrow x_k=x_{k+1}\)
\(\Rightarrow x_1=x_2=...=x_k=x_{k+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chắc là \(q\left(x\right)=x^2-4????\)
\(f\left(2\right)=2^5+2^2+1=37\) ; \(f\left(-2\right)=-27\)
Do \(f\left(x\right)\) có 5 nghiệm nên f(x) có dạng:
\(f\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)
\(\Rightarrow f\left(2\right)=\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)=37\)
\(f\left(-2\right)=\left(-2-x_1\right)\left(-2-x_2\right)\left(-2-x_3\right)\left(-2-x_4\right)\left(-2-x_5\right)=-27\)
\(\Rightarrow\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)=27\)
\(A=\left(x_1^2-4\right)\left(x^2_2-4\right)\left(x_3^2-4\right)\left(x_4^2-4\right)\left(x^2_5-4\right)\)
\(A=-\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)\)
\(A=-37.27=-999\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow2+\frac{x+4}{2000}+\frac{x+3}{2001}=2+\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2001}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
Suy ra x+2004=0
\(\Leftrightarrow x=-2004\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nhìn nó dài nhưng chỉ cần lập luận vài bước thui
Điều kiện : \(x_1,x_2,x_3,...,x_{2000}\ne0.\)
Từ (1) suy ra \(2x_1x_2=x_2^2+1>0\Rightarrow x_1\)và \(x_2\)cùng dấu.
Tương tự ta cũng có:
Từ (2) suy ra \(x_2\)và \(x_3\)cùng dấu
.....................................................
Từ (1999) suy ra \(x_{1999}\)và \(x_{2000}\)cùng dấu
Từ (2000) suy ra \(x_{2000}\)và \(x_1\)cùng dấu
Như vậy : các ẩn số \(x_1,x_2,...,x_{2000}\)cùng dấu .
Mặt khác nếu \(\left(x_1,x_2,...,x_{2000}\right)\)là một nghiệm thì \(\left(-x_1,-x_2,...,-x_{2000}\right)\)cũng là nghiệm . Do đó chỉ cần xét \(x_1,x_2,...,x_{2000}>0\).
Khi đó : \(2x_1=x_2+\frac{1}{x_2}\ge2\Rightarrow x_1\ge1\Rightarrow\frac{1}{x_1}\le1\)
\(2x_2=x_3+\frac{1}{x_3}\ge2\Rightarrow x_2\ge1\Rightarrow\frac{1}{x_2}\le1\)
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Tương tự , ta có: \(x_{2000}\ge1\Rightarrow\frac{1}{x_{2000}}\le1\)
Suy ra : \(\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_{2000}}\le x_1+x_2+...+x_{2000}\)
Mặt khác; nếu cộng từng vế 2000 phương trình của hệ , ta có:
\(x_1+x_2+...+x_{2000}=\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_{2000}}\)
Dấu '=' xảy ra khi và chỉ khi \(x_1=x_2=...=x_{2000}=1\)
Tóm lại hệ đã cho có 2 nghiệm :
\(\left(x_1,x_2,...,x_{2000}\right)=\left(1;1;...;1\right),\left(-1;-1;...;-1\right).\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi i là đại diện cho các số từ 1 đến 2011
ĐKXĐ: \(a_i\ne0\left(i=1,2,3,..,2011\right)\)
Xét \(a_i=1\) Ta có: \(\frac{1}{a^{11}_i}=1>\frac{2011}{2048}\Rightarrow\frac{1}{x^{11}_1}+\frac{1}{x^{11}_2}+...+\frac{1}{x^{11}_{2011}}>\frac{2011}{2048}\left(loai\right)\)
Xét \(a_i\ge2\) Ta có: \(\frac{1}{a^{11}_i}\le\frac{1}{2048}\Rightarrow\frac{1}{x^{11}_1}+\frac{1}{x^{11}_2}+...+\frac{1}{x^{11}_{2011}}\le\frac{2011}{2048}\)
Dấu "=" xảy ra khi \(a_i=2\)
Thay vào ta có:
\(M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow2M-M=\left(1+\frac{1}{2}+...+\frac{1}{2^{2010}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)\)
\(\Rightarrow M=1-\frac{1}{2^{2011}}\)