\(\dfrac{x+1}{2021}+\dfrac{x+2}{2020}=\dfrac{x+3}{2019}+\dfrac{x+4}{2018}\)

=>\(\dfrac{x+1}{2021}+1+\dfrac{x+2}{2020}+1=\dfrac{x+3}{2019}+1+\dfrac{x+4}{2018}+1\)

=>\(\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}=\dfrac{x+2022}{2019}+\dfrac{x+2022}{2018}\)

=> (x+2022)(\(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}-\dfrac{1}{2018}\))=0

=>x+2022=0

=> x=-2022

\(x-4\sqrt{x-2}+1=0\)(Đk x>2)

⇔\(x-2-4\sqrt{x-2}+4-1=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-2\right)^2-1=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-3\right)\left(\sqrt{x-2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}-3=0\\\sqrt{x-2}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=3\\\sqrt{x-2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=9\\x-2=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)(thảo đk)

Vậy\(\left[{}\begin{matrix}x=11\\x=3\end{matrix}\right.\)là nghiệm của pt

ĐKXĐ: x≥2

x+1=\(4\sqrt{x-2}\) bình phương 2 vế ta đc:\(\left(x+1\right)^2=16\cdot\left(x-2\right)< =>x^2+2x+1=16x-32< =>x^2-14x+33=0\)

giải phương trình này ta đc:x1=11(nhận); x2=3(nhận)

vậy phương trình có 2 nghiệm: x1=11;x2=3

\(\dfrac{x-1}{x-3}>1\left(x\ne3\right)\)

\(\Leftrightarrow\dfrac{x-1-x+3}{x-3}>0\)

\(\Leftrightarrow2>0\)

Vậy \(S=\left\{2\right\}\)

-ĐKXĐ: \(x\ne3\)

\(\dfrac{x-1}{x-3}>1\)

\(\Leftrightarrow\dfrac{x-1}{x-3}-\dfrac{x-3}{x-3}>0\)

\(\Leftrightarrow\dfrac{x-1-x+3}{x-3}>0\)

\(\Leftrightarrow\dfrac{2}{x-3}>0\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\)

-Vậy tập nghiệm của BĐT là {x l x>3}

\(x+\sqrt{9-x^2}-x\sqrt{9-x^2}=3\left(-3\le x\le3\right)\)

\(\Leftrightarrow\sqrt{9-x^2}-x\sqrt{9-x^2}=3-x\\ \Leftrightarrow9-x^2+x^2\left(9-x^2\right)-2x\sqrt{\left(9-x^2\right)^2}=9-6x+x^2\\ \Leftrightarrow9+8x^2-x^4-2x\left(9-x^2\right)=x^2-6x+9\\ \Leftrightarrow-x^4+2x^3+7x^2-12x=0\\ \Leftrightarrow-x\left(x^3-2x^2-7x+12\right)=0\Leftrightarrow-x\left(x^3-3x^2+x^2-3x-4x+12\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x^2+x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=3\left(N\right)\\x^2+x-4=0\left(1\right)\end{matrix}\right.\)

 \(\Delta\left(1\right)=1-4\left(-4\right)=17>0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1-\sqrt{17}}{2}\left(N\right)\\x=\dfrac{-1+\sqrt{17}}{2}\left(N\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;3;\dfrac{-1-\sqrt{17}}{2};\dfrac{-1+\sqrt{17}}{2}\right\}\)

Tick ✔

Sửa đề: (x-15)/17

=>\(\left(\dfrac{x-15}{17}-5\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-76}{12}-2\right)=0\)=>x-100=0

=>x=100

\(\left(x-1\right)\left(x+5\right)=x^2+4x-5\)(1)

\(x\left(x+4\right)=x^2+4x\)(2)

Lấy (1) nhân (2) \(\Leftrightarrow y.\left(y-5\right)=84\Leftrightarrow y^2-5y+\left(\frac{5}{2}\right)^2=84+\frac{25}{4}=\left(\frac{19}{2}\right)^2\)

\(\orbr{\begin{cases}y=\frac{5-19}{2}=-7\left(loai\right)\\y=\frac{5+19}{2}=12\end{cases}}\) 

\(x^2+4x=12\Leftrightarrow\left(x+2\right)^2=16\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)

`x(4x-4)-32>4x(x+1)`

`<=>4x^2-4x-32>4x^2+4x`

`<=>8x<-32`

`<=>x<-4`

Vậy `S={x|x<-4}`