\(\sqrt{x_1^2-1^2}+2\sqrt{x^2_2-2^2}+...+100\sqrt{x_{100}^2-100^2}=\dfrac{1}{2}\left(x_1^2+x^2_2+...+x_{100}^2\right)\)

\(\Leftrightarrow2\sqrt{x_1^2-1^2}+4\sqrt{x^2_2-2^2}+...+200\sqrt{x_{100}^2-100^2}=x_1^2+x^2_2+...+x_{100}^2\)

\(\Leftrightarrow x_1^2-1-2\sqrt{x_1^2-1}+1+x^2_2-4-4\sqrt{x^2_2-4}+4+...+x^2_{100}-10000-200\sqrt{x_{100}^2-10000}+10000=0\)

\(\Leftrightarrow\left(\sqrt{x^2_1-1}-1\right)^2+\left(\sqrt{x^2_2-4}-2\right)^2+....+\left(\sqrt{x^2_{100}-10000}-100\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2_1-1}-1=0\\\sqrt{x^2_2-4}-2=0\\....\\\sqrt{x^2_{100}-10000}-100=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\sqrt{1^2+1}=\sqrt{2}\\x_2=\sqrt{2^2+4}=2\sqrt{2}\\....\\x_{100}=\sqrt{100^2+10000}=100\sqrt{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x_1+x_2+...+x_{2000}=a\left(1\right)\\x_1^2+x_2^2+...+x_{2000}^2=a^2\left(2\right)\\x_1^{2000}+x_2^{2000}+...+x_{2000}^{2000}=a^{2000}\left(3\right)\end{matrix}\right.\)

Từ (2)(3)\(\Rightarrow2\left(x_1x_2+x_2x_3+...+x_{2000}x_1\right)=0\)

\(\Rightarrow x_1=x_2=...=x_{2000}=0\)

Vậy hpt có nghiệm là x=0.

Đúng không ạ?

a, \(\Delta'=\left(m-1\right)^2-\left(-2m+5\right)=m^2-2m+1+2m-5=m^2-4\)

Để pt vô nghiệm thì \(m^2-4< 0\Leftrightarrow-2< m< 2\)

Để pt có nghiệm kép thì \(m^2-4=0\Leftrightarrow m=\pm2\)

Để pt có 2 nghiệm phân biệt thì \(m^2-4>0\Leftrightarrow\left[{}\begin{matrix}m< -2\\m>2\end{matrix}\right.\)

2, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1x_2=-2m+5\end{matrix}\right.\)

\(a,ĐKXĐ:x_1,x_2\ne0\\ \dfrac{x_1}{x_2}+\dfrac{x_2}{x_1}=2\\ \Leftrightarrow\dfrac{x_1^2+x_2^2}{x_1x_2}=2\\ \Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=0\\ \Leftrightarrow\left(2m-2\right)^2-4\left(-2m+5\right)=0\\ \Leftrightarrow4m^2-8m+4+8m-20=0\\ \Leftrightarrow4m^2-16=0\\ \Leftrightarrow m=\pm2\)

\(b,x_1+x_2+2x_1x_2\le6\\ \Leftrightarrow2m-2+2\left(-2m+5\right)\le6\\ \Leftrightarrow2m-2-4m+10-6\le0\\ \Leftrightarrow-2m+2\le0\\ \Leftrightarrow m\ge1\)

Theo vi-et thì ta có:

\(\hept{\begin{cases}x_1+x_2=\frac{3a-1}{2}\\x_1x_2=-1\end{cases}}\)

Từ đây ta có: 

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=\left(\frac{3a-1}{2}\right)^2-4.1=\left(\frac{3a-1}{2}\right)^2-4\)

Theo đề bài thì 

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

\(=\frac{3}{2}.\left(x_1-x_2\right)^2+2.\left(x_1-x_2\right)^2\left(\frac{1}{2}-\frac{1}{x_1x_2}\right)^2\)

\(=\left(x_1-x_2\right)^2\left(\frac{3}{2}+2.\left(\frac{1}{2}-\frac{1}{x_1x_2}\right)^2\right)\)

\(=\left(\left(\frac{3a-1}{2}\right)^2-4\right)\left(\frac{3}{2}+2.\left(\frac{1}{2}+1\right)^2\right)\)

\(=6\left(\left(\frac{3a-1}{2}\right)^2-4\right)\ge6.4=24\)

Dấu = xảy ra khi \(a=\frac{1}{3}\)

Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1x_2=-5\end{matrix}\right.\)

Ta có : \(A=\left(x_1+x_2\right)^2-3x_1x_2\)

\(\Rightarrow4-3\left(-5\right)=4+15=19\)

Vậy A = 19 

ta thấy pt luôn có n_o . Theo hệ thức Vi - ét ta có:

x₁ + x₂ = \(\dfrac{-b}{a}\) = 6

x₁x₂ = \(\dfrac{c}{a}\) = 1

a) Đặt A = x₁\(\sqrt{x_1}\) + x₂\(\sqrt{x_2}\) = \(\sqrt{x_1x_2}\)( \(\sqrt{x_1}\) + \(\sqrt{x_2}\) )

=> A² = x₁x₂(x₁ + 2\(\sqrt{x_1x_2}\) + x₂)

=> A² = 1(6 + 2) = 8

=> A = 2\(\sqrt{3}\)

b) bạn sai đề

1. Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{4}{3}\\x_1.x_2=\dfrac{1}{3}\end{matrix}\right.\)

\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_1-1\right)\left(x_2-1\right)}\)

   \(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_1-x_2+1}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}\)

  \(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}=\dfrac{\dfrac{22}{9}}{\dfrac{8}{3}}=\dfrac{11}{12}\)

\(1,3x^2+4x+1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}=-\dfrac{4}{3}\\P=x_1x_2=\dfrac{c}{a}=\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(C=\dfrac{x_1}{x_2-1}+\dfrac{x_2}{x_1-1}\)

\(=\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{\left(x_2-1\right)\left(x_1-1\right)}\)

\(=\dfrac{x_1^2-x_1+x_2^2-x_2}{x_1x_2-x_2-x_1+1}\)

\(=\dfrac{\left(x_1^2+x_2^2\right)-\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{S^2-2P-S}{P-S+1}\)

\(=\dfrac{\left(-\dfrac{4}{3}\right)^2-2.\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)}{\dfrac{1}{3}-\left(-\dfrac{4}{3}\right)+1}\)

\(=\dfrac{11}{12}\)

Vậy \(C=\dfrac{11}{12}\)

\(3,3x^2-7x-1=0\)

Do pt có 2 nghiệm \(x_1,x_2\) nên theo đ/l Vi-ét ta có :

\(\left\{{}\begin{matrix}S=x_1+x_2=-\dfrac{b}{a}=\dfrac{7}{3}\\P=x_1x_2=\dfrac{c}{a}=-\dfrac{1}{3}\end{matrix}\right.\)

Ta có :

\(B=\dfrac{2x_2^2}{x_1+x_2}+2x_1\)

\(=\dfrac{2x_2^2+2x_1\left(x_1+x_2\right)}{x_1+x_2}\)

\(=\dfrac{2x_2^2+2x_1^2+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(x_1^2+x_2^2\right)+2x_1x_2}{x_1+x_2}\)

\(=\dfrac{2\left(S^2-2P\right)+2P}{S}\)

\(=\dfrac{2\left(\dfrac{7}{3}^2-2\left(-\dfrac{1}{3}\right)\right)+2\left(-\dfrac{1}{3}\right)}{\dfrac{7}{3}}\)

\(=\dfrac{104}{21}\)

Vậy \(B=\dfrac{104}{21}\)

Ta có: \(k\sqrt{x_k-k^2}\le\dfrac{1}{2}\left(k^2+x_k-k^2\right)=\dfrac{1}{2}x_k\)

\(\Rightarrow\sum\limits^{2005}_{k=1}k.\sqrt{x_k-k^2}\le\dfrac{1}{2}\left(x_1+x_2+...+x_{2005}\right)\)

Dấu "=" xảy ra khi:

\(k=\sqrt{x_k-k^2}\Leftrightarrow x_k=2k^2\) hay \(\left\{{}\begin{matrix}x_1=2.1^2=1\\x_2=2.2^2=8\\....\\x_{2005}=2.2005^2\end{matrix}\right.\)

Theo vi ét: \(\left\{{}\begin{matrix}x_1+x_2=6\\x_1x_2=8\end{matrix}\right.\)

Theo đề:

\(B=\dfrac{x_1\sqrt{x_1}-x_2\sqrt{x_2}}{x_1-x_2}=\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)\left(x_1+\sqrt{x_1x_2}+x_2\right)}{\left(\sqrt{x_1}-\sqrt{x_2}\right)\left(\sqrt{x_1}+\sqrt{x_2}\right)}\left(x_1,x_2\ge0\right)\)

\(=\dfrac{6+\sqrt{8}}{\sqrt{x_1}+\sqrt{x_2}}\)

Tính: \(\left(\sqrt{x_1}+\sqrt{x_2}\right)^2=x_1+x_2+2\sqrt{x_1x_2}=6+2\sqrt{8}=6+4\sqrt{2}=\left(\sqrt{4}+\sqrt{2}\right)^2\)

\(\Rightarrow\sqrt{x_1}+\sqrt{x_2}=\sqrt{4}+\sqrt{2}\) (thỏa mãn \(x_1,x_2\ge0\))

Khi đó: \(P=\dfrac{6+\sqrt{8}}{\sqrt{4}+\sqrt{2}}=4-\sqrt{2}\)