Question

[nguyễn ngọc quỳnh anh]

1 Tìm x:

a) (x-3)^3=27

b) x^3=x^6

c) (2x+2)^5=(2x+2)^6.

Answer 1

[FL.Han_]

a) \(\left(x-3\right)^3=27\)

\(\Leftrightarrow x-3=3\)

\(\Rightarrow x=6\)

b) \(x^3=x^6\)

\(\Leftrightarrow x^6-x^3=0\)

\(\Leftrightarrow x^3\left(x^3-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

c) \(\left(2x+2\right)^5=\left(2x+2\right)^6\)

\(\Leftrightarrow\left(2x+2\right)^5\left(2x+2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(2x+2\right)^5=0\\2x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-\frac{1}{2}\end{cases}}\)