Question

Bài 1 : Tìm x :

( x+1) ^x+2 = (x+1) ^x+4  

Các bn giúp mik với !!!!!!!!!!   Mik cảm ơn !

 

Answer 1

\(\left(x+1\right)^{x+2}=\left(x+1\right)^{x+4}\)

\(\Rightarrow x+2=x+4\)

\(\Rightarrow0x=2\)

=> không có giá trị của x thỏa mãn

=.= hk tốt!!

Answer 2

(x + 1)^{x + 2} = (x + 1)^{x + 4}

<=> x + 2 = x + 4

<=> 2 = x + 4 - x

<=> 2 = 4

<=> 0 = 4 - 2

<=> 0 = 2

=> không có x thỏa mãn đề bài

Answer 3

Ta có: \(\left(x+1\right)^{x+2}=\left(x+1\right)^{x+4}\)

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

=> \(\left(x+1\right)^{x+2}\left[1-\left(x+1\right)^2\right]=0\)

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

=> \(\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2=1\end{cases}}\)

=> x = -1

hoặc \(\orbr{\begin{cases}x+1=1\\x+1=-1\end{cases}}\) => \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

vậy ...

Answer 4

\(\left(x+1\right)^{x+2}=\left(x+1\right)^{x+4}\)

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

\(\Rightarrow\left(x+1\right)^{x+2}\left[\left(x+1\right)^2-1\right]=0\)

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

\(\Rightarrow x+1=0\)\(\Rightarrow x=-1\)

\(\Rightarrow x+1=1\)\(\Rightarrow x=0\)

\(\Rightarrow x+1=-1\)\(\Rightarrow x=-2\)

Vậy\(x=-1;0;-2\)