Question

Tìm x biết:

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

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

Answer 1

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

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

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

Trường hợp 1 : 

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

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

\(\Leftrightarrow\)\(x-3=0\)

\(\Leftrightarrow\)\(x=3\)

Trường hợp 2 : 

\(1-\left(x-3\right)^{10}=0\)

\(\Leftrightarrow\)\(\left(x-3\right)^{10}=1\)

\(\Leftrightarrow\)\(\left(x-3\right)^{10}=1^{10}\)

\(\Leftrightarrow\)\(x-3=1\)

\(\Leftrightarrow\)\(x=4\)

Vậy \(x=3\) hoặc \(x=4\)

Chúc bạn học tốt ~