x(x+1)-(x-2)(x+1)=0

\(\left(x+1\right)\left(x-x+2\right)=0\\ \left(x+1\right)\cdot2=0\\ =>x+1=0\\ x=0-1\\ x=-1\)

=>(x+1)(x-x+2)=0

=>x+1=0

=>x=-1

\(\Leftrightarrow\left(x-2021\right)\left(x-5\right)-\left(x-2021\right)=0\\ \Leftrightarrow\left(x-2021\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2021\\x=6\end{matrix}\right.\)

a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)

ĐK:\(x\ge0\)

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

Ủa lớp 7 sao học căn r nè

\(\dfrac{1}{3}x+\dfrac{2}{3}\left(x-1\right)=0\\ \dfrac{1}{3}x+\dfrac{2}{3}x-\dfrac{2}{3}=0\\ x=\dfrac{2}{3}\)

`1/3x + 2/3(x-1) =0`

` 1/3x + 2/3x -2/3 = 0`

` ( 1/3 + 2/3) x -2/3 = 0`

` 3/3x -2/3 = 0`

` 1x-2/3 = 0`

`1/x = 0 + 2/3`

` 1x = 2/3`

` x = 2/3`

Ta có: \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+x\left(7x-6\right)=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+7x^2-6x=0\)

\(\Leftrightarrow x^2+7x-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)

bạn có thể tách rõ hơn đoạn cuối dc khum mình cảm ơn

à k cần nữa ạ

a: (x-1)(x+2)(-x-3)=0

=>(x-1)(x+2)(x+3)=0

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

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

b: (x-7)(x+3)<0

TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)

=>-3<x<7

mà x nguyên

nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)

đánh đề bằng latex cho rõ đi bạn, không biết nào dấu nào biến:v

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

\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)