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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
=> \(x_1+\frac{1}{x_1}=x_2+\frac{1}{x_2}=x_3+\frac{1}{x_3}=...x_{k+1}+\frac{1}{x_1}\)
=> \(x_1+\frac{1}{x_2}=x_2+\frac{1}{x_3}=....=x_k+\frac{1}{x_{k+1}}=x_{k+1}+\frac{1}{x_1}\)
\(\Rightarrow x_{k-1}+\frac{1}{x_k}=x_k+\frac{1}{x_1}=x_{k+1}+\frac{1}{x_1}\)
\(\Rightarrow x_k-x_{k+1}=0\)
\(\Rightarrow x_k=x_{k+1}\)
\(\Rightarrow x_1=x_2=...=x_k=x_{k+1}\)
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:
\(f\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)
\(\Rightarrow f\left(2\right)=\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)=37\)
\(f\left(-2\right)=\left(-2-x_1\right)\left(-2-x_2\right)\left(-2-x_3\right)\left(-2-x_4\right)\left(-2-x_5\right)=-27\)
\(\Rightarrow\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)=27\)
\(A=\left(x_1^2-4\right)\left(x^2_2-4\right)\left(x_3^2-4\right)\left(x_4^2-4\right)\left(x^2_5-4\right)\)
\(A=-\left(2-x_1\right)\left(2-x_2\right)\left(2-x_3\right)\left(2-x_4\right)\left(2-x_5\right)\left(2+x_1\right)\left(2+x_2\right)\left(2+x_3\right)\left(2+x_4\right)\left(2+x_5\right)\)
\(A=-37.27=-999\)
BĐT Cauchy-Schwarz:
\(\left(1+1+1+...+1\right)\left(x^2_1+x^2_2+...+x^2_{2017}\right)\ge\left(x_1+x_2+...+x_{2017}\right)^2\left(\text{2017 số 1}\right)\)
\(\Leftrightarrow2017\left(x^2_1+x^2_2+...+x^2_{2017}\right)\ge\left(x_1+x_2+...+x_{2017}\right)^2\)
\(\Leftrightarrow x^2_1+x^2_2+...+x^2_{2017}\ge\dfrac{\left(x_1+x_2+...+x_{2017}\right)^2}{2017}\)
Khi \(\dfrac{x_1}{1}=\dfrac{x_2}{1}=...=\dfrac{x_{2017}}{1}\Leftrightarrow x_1=x_2=...=x_{2017}\)
Bạn j j biết làm bài ơi, giải hộ với. Bạn chưa biết làm thì nghĩ hộ t với. Làm được tớ cho mấy cái kẹo mút này...
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:
\(M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow2M-M=\left(1+\frac{1}{2}+...+\frac{1}{2^{2010}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)\)
\(\Rightarrow M=1-\frac{1}{2^{2011}}\)