Question

Cho mình xin hỏi 

(X+3)^5=-32

(2x+8).(x^2-4)=0

Answer 1

Lớp 5 chưa học lũy thừa và số nguyên âm nhé bạn :D

\((x+3)^5=-32\)

\(\Rightarrow(x+3)^5=(-2)^5\)

\(\Rightarrow x+3=-2\)

\(\Rightarrow x=-2-3\Rightarrow x=-5\)

\((2x+8)\cdot(x^2-4)=0\)

Làm nốt

P/S : Hoq chắc :>

Answer 2

\(\left(x+3\right)^5=-32\)

\(\Rightarrow\left(x+3\right)^5=\left(-2\right)^5\)

\(\Rightarrow x+3=-2\)

\(\Rightarrow x=-2-3=-5\)

Answer 3

\(\left(2x+8\right).\left(x^2-4\right)=0\)

\(\Rightarrow\hept{\begin{cases}2x+8=0\\x^2-4=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=0-8\\x^2=0+4\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=-8\\x^2=4\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-4\\x=2\\x=-2\end{cases}}\)

Answer 4

*)Ta có:

(x+3)⁵=-32

\(\Rightarrow\)(x+3)⁵=(-2)⁵

\(\Rightarrow\)x+3=-2

\(\Rightarrow\)x    =(-2)-3

\(\Rightarrow\)x   =-5

*)Ta có:

(2x+8)(x²-4)=0

\(\Rightarrow\orbr{\begin{cases}2x+8=0\\x^2-4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=-8\\x^2=4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-4\\x\in\left\{2;-2\right\}\end{cases}}\)

Vậy \(x\in\left\{-4;-2;2\right\}\)

Answer 5

1.(x+3)^5=-32                               2. TH1 2x+8=0 =>x=-4

           =-2^5                                    TH2 x^2-4=0 =>x=2

 =>x+3=-2

     x    =-2-3

     x    =-5

Answer 6

\((2x+8)(x^2-4)=0\)

\(\Rightarrow\hept{\begin{cases}2x+8=0\\x^2-4=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=0-8\\x^2=0+4\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=-8\\x^2=4\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=-4\\x\pm2\end{cases}}\)

Vậy :

P/S : Hoq chắc :>