(x-2)(x+3)=0 (2x-3)(x^2+1)=0 3x(x-2)+4(x-2)=0 (3x-4)^2-(x-3)^2=0

Đáp án:

`↓`

Giải thích các bước giải:

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

`=>`  \(\left[ \begin{array}{l}x- 2 = 0\\x + 3 = 0\end{array} \right.\)

`=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 

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

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

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

`=>` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\ \text{( lấy )} \\ x=-1\ \text{( loại )}\end{array} \right.\) 

`3x(x-2) + 4(x-2) = 0`

`=> ( 3x + 4 )( x - 2 ) = 0`

`=>` \(\left[ \begin{array}{l}3x + 4 = 0\\x - 2 = 0\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x = \dfrac{-4}{3}\\x=2\end{array} \right.\) 

`( 3x - 4 )^2 - ( x - 3 )^2 = 0`

`=> (3x -  4 + x - 3)(3x - 4 - x + 3) = 0`

`=>  (4x - 7)(2x - 1) = 0`

`=>` \(\left[ \begin{array}{l}4x - 7 = 0\\2x - 1 = 0\end{array} \right.\)

`=>`  \(\left[ \begin{array}{l}x=\dfrac{7}{4}\\x = \dfrac{1}{2}\end{array} \right.\) 

`#Sad`

Đáp án+Giải thích các bước giải:

$(x-2)(x+3)=0^{}$

`=>` \(\left[ \begin{array}{l}x-2=0\\x+3=0\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 

Vậy $x=2$ hoặc $x=-3$

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$(2x-3)(x^2+1)=0^{}$

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

`=>` \(\left[ \begin{array}{l}2x=3\\x^2=-1\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x^2=-1 (ko có giá trị của x)\end{array} \right.\) 

Vậy $x=$$\dfrac{3}{2}$

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$3x(x-2)+4(x-2)=0^{}$

`=>` $(x-2)(3x+4)=0^{}$

`=>` \(\left[ \begin{array}{l}x-2=0\\3x+4=0\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=2\\3x=-4\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=2\\x=\dfrac{-4}{3}\end{array} \right.\) 

Vậy $x=2$ hoặc $x=$$\dfrac{-4}{3}$

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$(3x-4)^2-(x-3)^2=0^{}$

`=>` $[(3x-4)+(x-3)].[(3x-4-(x-3)]=0^{}$

`=>` $(3x-4+x-3)(3x-4-x+3)=0^{}$

`=>` $(4x-7)(2x-1)=0^{}$

`=>` \(\left[ \begin{array}{l}4x-7=0\\2x-1=0\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}4x=7\\2x=1\end{array} \right.\) 

`=>` \(\left[ \begin{array}{l}x=\dfrac{7}{4}\\x=\dfrac{1}{2}\end{array} \right.\) 

Vậy $x=$$\dfrac{7}{4}$ hoặc $x=$$\dfrac{1}{2}$