Đáp án+Giải thích các bước giải:

 `1)`

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

`→x^2+x-3=2x^2-2x-x+1`

`→x^2+x-3=2x^2-3x+1`

`→2x^2-3x+1-x^2-x+3=0`

`→x^2-4x+4=0`

`→x^2-2.x.2+2^2=0`

`→(x-2)^2=0`

`→x-2=0`

`→x=2`

Vậy `x=2`

`2)`

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

`→3x^2+x+6x+2-x^2+4=0`

`→2x^2+7x+6=0`

`→2x^2+4x+3x+6=0`

`→2x(x+2)+3(x+2)=0`

`→(x+2)(2x+3)=0`

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

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

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

Vậy `x∈\{-2;-3/2\}`

`3)`

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

`→x^2-3x+2=x-x^2+2x-2`

`→x^2-3x+2=-x^2+3x-2`

`→x^2-3x+2+x^2-3x+2=0`

`→2x^2-6x+4=0`

`→x^2-3x+2=0`

`→x^2-x-2x+2=0`

`→x(x-1)-2(x-1)=0`

`→(x-1)(x-2)=0`

\(→\left[ \begin{array}{l}x-1=0\\x-2=0\end{array} \right.\) 

\(→\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\)

Vậy `x∈\{1;2\}`

Đáp án:

Giải thích các bước giải:

`1, x( x + 1 ) - 3 = ( 2x - 1 )( x - 1 )`

`<=> x^2 + x - 3 = 2x^2 - 2x - x + 1`

`<=> x^2 - 2x^2 + x + 2x + x - 3 - 1 = 0`

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

`<=> -x^2 + 4x - 4 = 0`

`<=> -x^2 + 2x + 2x - 4 = 0`

`<=> ( -x^2 + 2x ) + ( 2x - 4 ) = 0`

`<=> -x ( x - 2 ) + 2 ( x - 2 ) = 0`

`<=> ( -x + 2 )( x - 2 ) = 0`

`<=> - ( x - 2 )( x - 2 ) = 0`

`<=> ( x - 2 )^2 = 0`

`<=> x - 2 = 0`

`<=> x = 2 `

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

`<=> 3x^2 + x + 6x + 2 - x^2 + 4 = 0`

`<=> ( 3x^2 - x^2 ) + ( x + 6x ) + ( 2 + 4 ) = 0`

`<=> 2x^2 + 7x + 6 = 0`

`<=> 2x^2 + 2x + 3/2x + 6 = 0`

`<=> ( 2x^2 + 4x ) + ( 3x + 6 ) = 0`

`<=> 2x( x + 2 ) + 3 ( x + 2 ) = 0`

`<=> ( x + 2 )( 2x + 3 ) = 0`

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

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

`3, x( x - 3 ) + 2 = x( 1 - x ) + 2x - 2`

`<=> x^2 - 3x + 2 = x - x^2 + 2x - 2`

`<=> x^2 - 3x + 2 - x + x^2 - 2x + 2 = 0`

`<=> ( x^2 + x^2 ) - ( 3x + x + 2x ) + ( 2 + 2 ) = 0`

`<=> 2x^2 - 6x + 4 = 0`

`<=> 2 ( x^2 - 3x + 2 ) = 0`

`<=> x^2 - x - 2x + 2 = 0`

`<=> ( x^2 - x ) - ( 2x - 2 ) = 0`

`<=> x ( x - 1 ) - 2 ( x - 1 ) = 0`

`<=> ( x - 2 )( x - 1 ) = 0`

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

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