Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(ĐK:x\ne\pm2;x\ne-1\\ PT\Leftrightarrow\dfrac{x^2-x-2+x^2+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x+1}{x+1}\\ \Leftrightarrow\dfrac{2x^2-4}{x^2-4}=\dfrac{2x+1}{x+1}\\ \Leftrightarrow\left(2x^2-4\right)\left(x+1\right)=\left(x^2-4\right)\left(2x+1\right)\\ \Leftrightarrow2x^3+2x^2-4x-4=2x^3+x^2-8x-4\\ \Leftrightarrow x^2+4x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-4\left(tm\right)\end{matrix}\right.\)
\(\dfrac{x-2}{x+1}-\dfrac{3}{x+2}>0.\left(x\ne-1;-2\right).\\ \Leftrightarrow\dfrac{x^2-4-3x-3}{\left(x+1\right)\left(x+2\right)}>0.\\ \Leftrightarrow\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)
Đặt \(f\left(x\right)=\dfrac{x^2-3x-7}{\left(x+1\right)\left(x+2\right)}>0.\)
Ta có: \(x^2-3x-7=0.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{37}}{2}.\\x=\dfrac{3-\sqrt{37}}{2}.\end{matrix}\right.\)
\(x+1=0.\Leftrightarrow x=-1.\\ x+2=0.\Leftrightarrow x=-2.\)
Bảng xét dấu:
\(\Rightarrow f\left(x\right)>0\Leftrightarrow x\in\left(-\infty-2\right)\cup\left(\dfrac{3-\sqrt{37}}{2};-1\right)\cup\left(\dfrac{3+\sqrt{37}}{2};+\infty\right).\)
\(\sqrt{x^2-3x+2}\ge3.\\ \Leftrightarrow x^2-3x+2\ge9.\\ \Leftrightarrow x^2-3x-7\ge0.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3-\sqrt{37}}{2}.\\x=\dfrac{3+\sqrt{37}}{2}.\end{matrix}\right.\)
Đặt \(f\left(x\right)=x^2-3x-7.\)
\(f\left(x\right)=x^2-3x-7.\)
\(\Rightarrow f\left(x\right)\ge0\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)
\(\Rightarrow\sqrt{x^2-3x+2}\ge3\Leftrightarrow x\in(-\infty;\dfrac{3-\sqrt{37}}{2}]\cup[\dfrac{3+\sqrt{37}}{2};+\infty).\)
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+y}+6x-3y=6\\\dfrac{3}{x+y}+2x-4y=1\end{matrix}\right.\)
\(\Rightarrow4x+y=5\Rightarrow y=5-4x\)
Thế vào phương trình đầu:
\(\dfrac{1}{x+5-4x}+2x-\left(5-4x\right)=2\)
\(\Leftrightarrow\dfrac{1}{5-3x}+6x-7=0\)
\(\Leftrightarrow\left(6x-7\right)\left(5-3x\right)+1=0\)
\(\Leftrightarrow...\)
\(\dfrac{-x^2-3x+18}{\left(x-2\right)\left(x+2\right)}< 0\)
\(\Leftrightarrow\dfrac{x^2+3x-18}{\left(x-2\right)\left(x+2\right)}>0\)
Theo BXD, ta có: f(x)>0
=>\(x\in\left(-\infty;-6\right)\cup\left(-2;2\right)\cup\left(3;+\infty\right)\)
\(1)\sqrt{x^2+1}< 3.\\ \Leftrightarrow x^2+1< 9.\\ \Leftrightarrow x^2< 8.\\ \Leftrightarrow\left[{}\begin{matrix}x< 2\sqrt{2}.\\x>-2\sqrt{2}.\end{matrix}\right.\)
\(\Leftrightarrow-2\sqrt{2}< x< 2\sqrt{2}.\)
\(2)\dfrac{x^2-4x+3}{x^2-4}< 0.\)
Đặt \(f\left(x\right)=\dfrac{x^2-4x+3}{x^2-4}.\)
\(x^2-4=0.\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=-2.\end{matrix}\right.\\ x^2-4x+3=0.\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=1.\end{matrix}\right.\)
Bảng xét dấu:
\(\Rightarrow f\left(x\right)< 0\Leftrightarrow x\in\left(-2;1\right)\cup\left(2;3\right).\)
Lời giải:
1.
$\sqrt{x^2+1}<3$
$\Leftrightarrow 0\leq x^2+1<9$
$\Leftrightarrow x^2+1<9$
$\Leftrightarrow x^2<8$
$\Leftrightarrow -2\sqrt{2}< x< 2\sqrt{2}$
2.
Xét 2 TH:
TH1: \(\left\{\begin{matrix} x^2-4x+3<0\\ x^2-4>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x-1)(x-3)<0\\ (x-2)(x+2)>0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 1< x< 3\\ x>2 \text{hoặc} x<-2\end{matrix}\right.\)
\(\Leftrightarrow 2< x<3\)
TH2: \(\left\{\begin{matrix} x^2-4x+3>0\\ x^2-4<0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x-1)(x-3)>0\\ (x-2)(x+2)<0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x>3 \text{hoặc} x<1\\ -2< x< 2\end{matrix}\right.\)
\(\Leftrightarrow -2< x< 1\)
Kết hợp 2 TH suy ra tập nghiệm \(S=(2;3)\cup (-2;1)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x-2>=0\\4-x>=0\\x+1< >0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2< =x< =4\\x< >-1\end{matrix}\right.\Leftrightarrow x\in\left[2;4\right]\)
Đặt \(\left(x;2y;4z\right)=\left(a;b;c\right)\Rightarrow a+b+c=2019\)
Ta cần chứng minh: \(\frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ac}{c+a}\le2019\)
Áp dụng BĐT \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\Rightarrow VT=\frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ac}{c+a}\le\frac{2ab}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{2bc}{4}\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{2ac}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(\Rightarrow VT\le\frac{b}{2}+\frac{a}{2}+\frac{c}{2}+\frac{b}{2}+\frac{a}{2}+\frac{c}{2}=a+b+c=2019\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{2019}{3}\) hay \(\left(x;y;z\right)=\left(\frac{2019}{3};\frac{2019}{6};\frac{2019}{12}\right)\)