Question

Tìm x :

a)\(\dfrac{x-1}{x=5}=\dfrac{6}{7}\)

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

Answer 1

Này bạn, hình như bạn viết sai dấu rồi, mình không hiểu\(\dfrac{x-1}{x=5}=\dfrac{6}{7}\)là gì???

Answer 2

bạn viết lại đề câu a đi!

Answer 3

a) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\Rightarrow7\left(x-1\right)=6\left(x+5\right)\)

\(\Rightarrow7x-7=6x+30\Rightarrow7x-6x=30+7\)

\(\Rightarrow x=37\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

Vậy...............

Answer 4

a)\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)

\(\Rightarrow6x+30=7x-7\)

\(\Rightarrow7+30=7x-6x\)

\(\Rightarrow x=37\)

Vậy x=37

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

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

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

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{1;2\right\}\)