Tìm x : a) ( x + 3 )^2 – x( x + 5 ) = 2 b) ( 5x – 2 )^2 + ( 2 – 5x ) ( 3x + 1 ) = 0 c) x^3 – ( x + 3 ) ( x ^2 – 3x + 9 ) + ( 3x – 1 ) ( 3x + 1 )

Đáp án:

a,`(x + 3)^2 – x(x + 5)=  2`

`<=> x^2 + 6x + 9 – x^2 – 5x = 2`

`<=> x + 9 = 2`

`<=> x = -7`

b, `(5x – 2)^2 + (2 – 5x)(3x + 1) = 0`

`<=> 25x^2 – 20x + 4 + 6x – 15x^2 + 2 – 5x = 0`

`<=> 10x^2 – 19x + 6 = 0`

`<=> 10x^2 – 15x – 4x + 6 = 0`

`<=> 5x.(2x – 3) – 2(2x – 3) = 0`

`<=> (5x – 2)(2x – 3) = 0`

<=> \(\left[ \begin{array}{l}5x – 2 = 0\\2x – 3 = 0\end{array} \right.\) 

<=> \(\left[ \begin{array}{l}x = 2/5\\x= 3/2\end{array} \right.\) 

Vậy `S = {2/5 ; 3/2}`

c, ` x^3 + 27 + ( x + 3 ) ( x – 9 ) = 0`

`<=> (x + 3)(x^2 – 3x + 9) + (x + 3)(x – 9) = 0`

`<=> (x + 3)(x^2 – 3x + 9 + x – 9) = 0`

`<=> (x + 3)(x^2 – 2x) = 0`

`<=> x(x + 3)(x – 2) = 0`

`<=> x = 0` hoặc `x + 3 = 0` hoặc `x – 2 = 0`

`<=> x = 0` hoặc `x = -3` hoặc `x = 2`

Vậy `S = {0 ; -3 ; 2}`

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