Question

Tìm x

\(\left(x+1\right).\left(x^2-4\right)=0\)

Answer 1

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x^2-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x^2=4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\left(-2\right)\end{cases}}\)

Vậy x = -1 hoặc x = 2 hoặc x = -2

Answer 2

Khó!!!

Answer 3

\(\left(x+1\right)\left(x^2-4\right)=0\)

\(=\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)

Th1 : \(x+1=0\Rightarrow x=-1\)

Th2 : \(x-2=0\Rightarrow x=2\)

Th3 : \(x+2=0\Rightarrow x=-2\)

Answer 4

\(\left(x+1\right)\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x^2=4\end{cases}}}\)

Ta có:\(x^2=4\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy \(x\in\left\{-2,-1,2\right\}\) thỏa mãn

Answer 5

ta thấy phải có 1 vế = 0 mà vế x + 1 ko thể bằng 0 được => \(x^2-4=0\)

vậy \(x^2=4\)

=> \(x=2\)

Answer 6

\(\left(x+1\right).\left(x^2-4\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)

TH1 \(x+1=0\Rightarrow x=-1\)

TH2\(x-2=0\Rightarrow x=2\)

TH3\(x+2=0\Rightarrow x=-2\)

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Answer 7

(x + 1) x 2 − 4 = 0 = x + 1 x − 2 x + 2 = 0

Th1 : x + 1 = 0⇒x = −1

2    : x − 2 = 0⇒

3     x = 2 Th3 : x + 2 =

0⇒x = −2