Question

Tìm x,y biết

a,   \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)

b,    \(\frac{1+3y}{12}=\frac{1+5y}{5x}=\frac{1+7y}{4x}\)

c,    \(\frac{1}{2018}:2017x=-\frac{1+7y}{4x}\)

Giúp mình nhé, thứ 4 nộp r

 

Answer 1

a) \(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+15}=0\)

\(\left(x-3\right)^{x+5}-\left(x-3\right)^{x+5}\cdot\left(x-3\right)^{10}=0\)

\(\left(x-3\right)^{x+5}\cdot\left[1-\left(x-3\right)^{10}\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^{x+5}=0\\1-\left(x-3\right)^{10}=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{10}=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\\left(x-3\right)^{10}=\left(\pm1\right)^{10}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=\left\{4;2\right\}\end{cases}}\)

Vậy........