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 )

0 bình luận về “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 )”

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}`

   Giải thích các bước giải:

    

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2. a, (x+3)²-x(x+5)=2

   ⇔ x²+2.x.3+3²-x²-5x=2

   ⇔ x²+6x+9-x²-5x=2

   ⇔ x+9=2

   ⇔ x=2-9

   ⇔ x= -7

   b, (5x-2)²+(2-5x)(3x+1)=0

   ⇔ (5x-2)² +2(3x+1)-5x(3x+1)=0

   ⇔ [(5x)²-2.5x.2+2²]+6x+2-15x²-5x=0

   ⇔ 25x²-20x+4+x+2-15x²=0

   ⇔ 10x²-19x+6=0

   ⇔ 10x²-15x-4x+6=0

   ⇔ 5x(2x-3)-4(2x-3)=0

   ⇔ (5x-4)(2x-3)=0

   ⇔ 5x-4=0 hoặc 2x-3=0

   ⇔ x=4/5 hoặc x=3/2

   c, x³-(x+3)(x²-3x+9)+(3x-1)(3x+1)

   = x³-(x³+3³)+[(3x)²-1²]

   = x³-x³-27+9x²-1

   = 9x²-28

   c, x³ + 27 + ( x + 3 ) ( x – 9 ) = 0

   ⇔ x³+27+x(x-9)+3(x-9)=0

   ⇔ x³+27+x²-9x+3x-27=0

   ⇔ x³+x²-6x=0
   

   ⇔ x³-2x²+3x²-6x=0

   ⇔ x²(x-2)+3x(x-2)=0

   ⇔ (x²+3x)(x-2)=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

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