Tìm x€Z : (x²-9).(3x+15)=0 (4x-8).(x²+1)=0 (x+1).(3-x)=0 (x-2).(2x-1)=0 (3x+9).(1-3x)=0 (x²+1).(81-x²)=0

Tìm x€Z :
(x²-9).(3x+15)=0
(4x-8).(x²+1)=0
(x+1).(3-x)=0
(x-2).(2x-1)=0
(3x+9).(1-3x)=0
(x²+1).(81-x²)=0

0 bình luận về “Tìm x€Z : (x²-9).(3x+15)=0 (4x-8).(x²+1)=0 (x+1).(3-x)=0 (x-2).(2x-1)=0 (3x+9).(1-3x)=0 (x²+1).(81-x²)=0”

  1. `a)(x^2-9)(3x+15)=0`

    `→` \(\left[ \begin{array}{l}x^2-9=0\\3x+15=0\end{array} \right.\) 

    `→` \(\left[ \begin{array}{l}x^2=9\\3x=-15\end{array} \right.\) 

    `→` \(\left[ \begin{array}{l}x=±3\\x=-5\end{array} \right.\)

    Vậy `x∈{3;-3;-5}`

    `b)(4x-8)(x^2+1)=0`

    `→4x-8=0` (vì `x^2+1≥1`)

    `→4x=8`

    `→x=2`

    Vậy `x=2`

    `c)(x+1)(3-x)=0`

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

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

    Vậy `x∈{-1;3}`

    `d)(x-2)(2x-1)=0`

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

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

    Vậy `x∈{2;1/2}`

    `e)(3x+9)(1-3x)=0`

    `→` \(\left[ \begin{array}{l}3x+9=0\\1-3x=0\end{array} \right.\) 

    `→` \(\left[ \begin{array}{l}x=-3\\x=\dfrac{1}{3}\end{array} \right.\) 

    Vậy `x∈{-3;1/3}`

    `f)(x^2+1)(81-x^2)=0`

    `→81-x^2=0` (vì `x^2+1≥1`)

    `→x^2=81`

    `→x^2=9^2`

    `→` \(\left[ \begin{array}{l}x=9\\x=-9\end{array} \right.\) 

    Vậy `x∈{9;-9}`

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  2. (x²-9).(3x+15)=0

    => x² – 9 = 0          hoặc 3x + 15 =0

    Xét x² – 9 = 0

    => x² = 9

    => x = ±3 

    Xét 3x + 15 = 0

    => 3x = -15

    => x = -5

    (4x-8).(x²+1)=0

    => 4x – 8 = 0              hoặc x² – 1 =0

    Xét : 4x – 8 = 0

    => 4x = 8

    => x = 2

    Xét x² – 1 = 0

    => x² = 1

    => x = 1

    (x+1).(3-x)=0

    => x + 1 =0     hoặc   3 – x = 0

    Xét x + 1 = 0

    => x = -1

    Xét : 3 – x = 0

    => -x = -3 

    => x = 3

    (x-2).(2x-1)=0

    => x – 2 = 0  hoặc 2x – 1 = 0

    Xét : x – 2 = 0

    => x = 2

    Xét 2x – 1 = 0 

    => 2x = 1 

    => x = 0,5

    (3x+9).(1-3x)=0

    => 3x + 9 = 0 hoặc 1 – 3x = 0

    Xét 3x + 9 = 0

    => 3x = -9 

    => x = -3

    Xét : 1 – 3x = 0

    => 3x = 1

    => x = $\frac{1}{3}$ 

    (x²+1).(81-x²)=0

     => x² + 1 = 0    hoặc 81 – x² = 0

    Xét : x² + 1 =0 

    => x² = -1 (Vô lý )

    Xét 81 – x² = 0

    => x² = 81

    => x = ± 9

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