\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+5x\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

a: =>3x+3=4x-4

=>-x=-7

hay x=7(nhận)

b: (x-1)(x-3)=0

=>x-1=0 hoặc x-3=0

=>x=1 hoặc x=3

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

=>2x-2+x=0

=>3x-2=0

hay x=2/3

a, ĐKXĐ : x ≠ 1 ; x ≠ -1

\(\Rightarrow3\left(x+1\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x+3=4x-4\)

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\left(N\right)\)

b,

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

c,

\(\Leftrightarrow2x-2+x=0\)

\(\Leftrightarrow3x=2\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

\(\left(3x-1\right)^2.\left(x+5\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

^2: Nếu trong ngoặc bằng 0 thì bình phương cũng bằng 0 và ngược lại

\(x^2+2y^2-2xy+4y+3< 0\)

\(\Rightarrow x^2-2xy+y^2+y^2+4y+4-1< 0\)  

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)-1< 0\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Mà: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\) 

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1\ge-1\forall x,y\)

Mặt khác: \(\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Dấu "=" xảy ra:

\(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\)

\(\Rightarrow x=y=-2\)

Vậy: .... 

bn ko nói là tìm x hay y hay xy à

`xy-x-y=0`

`<=>xy-x-y+1=1`

`<=> x(y-1)-(y-1)=1`

`<=> (y-1)(x-1)=1`

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

\(\Leftrightarrow2\left(x^2-\dfrac{3}{2}x+\dfrac{5}{2}\right)=0\\ \Leftrightarrow2\left(x-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{31}{16}\right)=0\\ \Leftrightarrow2\left(x-\dfrac{3}{4}\right)^2+\dfrac{31}{8}=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-\dfrac{3}{4}\right)^2+\dfrac{31}{8}\ge\dfrac{31}{8}>0\right]\)

\(\left(x-3\right)\left(x^2+5\right)=0\)

\(\Rightarrow x-3=0\) hoặc \(x^2+5=0\)

      x= 3                   \(x^2=-5\) ( vô lý- do \(x^2\ge0\) với mọi x)

Vậy ...

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)