Tìm X: a)(x+1)(x^2-4)=0 b) (x-2) . (x^2+1)=0 c) 13.(x-5)= -169 d)x.(x-2)=0

2 bình luận về “Tìm X: a)(x+1)(x^2-4)=0 b) (x-2) . (x^2+1)=0 c) 13.(x-5)= -169 d)x.(x-2)=0”

1. a, (x + 1)(x^2 – 4) = 0

    TH1:

   x + 1 = 0

   => x = 0 – 1

   => x = -1

   TH2:

   x^2 – 4 = 0

   => x^2 = 0 + 4

   => x^2 = 4

   => x^2 = 2^2 = (-2)^2

   => $\left[\begin{matrix} x = 2\\ x = -2\end{matrix}\right.$

   Vậy x in {-1 ; 2 ; -2}

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   b, (x – 2)(x^2 + 1) = 0

   TH1:

   x – 2 = 0

   => x = 2

   TH2:

   x^2 + 1 = 0

   => x^2 = -1

   => Không có giá trị thỏa mãn x

   Vậy x = 2

   ______________________________________________

   c) 13(x – 5) = -169

   =>x – 5 = -169 : 13

   => x – 5 = -13

   => x = -13 + 5

   => x = -8

   Vậy x = -8

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   d) x(x – 2) = 0

   TH1:

   x = 0

   TH2:

   x – 2 = 0

   => x = 2

   Vậy x in {0 ; 2}

   text{#maingoctranthi}

   Trả lời
2. a) (x+1)($x^{2}-4$)=0

   <=> $\left \{ {{x+1=0} \atop {x^{2}=4}} \right.$ 

   <=> $\left \{ {{x+1=0} \atop {x=±2}} \right.$ 

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

   mà $x^{2}+1$>0

   => x-2 = 0 <=> x = 2

   c) 13(x-5)=-169

   <=> x-5=-13

   <=> x = -8

   d) x(x-2)=0

   <=> $\left \{ {{x=0} \atop {x-2=0}} \right.$ 

   <=> $\left \{ {{x=0} \atop {x=2}} \right.$ 

   Trả lời