Question

giúp e với ạ

Answer 1

a, đk : x >= 6 

\(\left[{}\begin{matrix}2x=x-6\\2x=6-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\left(ktm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

vậy pt vô nghiệm 

b, đk x >= -16/3 

\(\left[{}\begin{matrix}-5x-16=3x\\-5x=-3x-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=8\left(tm\right)\end{matrix}\right.\)

c, tương tự 

d, \(\left|3-x\right|+x^2-4x-x^2=0\Leftrightarrow\left|3-x\right|=4x\)đk >= 0 

\(\left[{}\begin{matrix}3-x=4x\\3-x=-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(ktn\right)\end{matrix}\right.\)

e, \(\left[{}\begin{matrix}4-5x=5-6x\\4-5x=6x-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{9}{11}\end{matrix}\right.\)

f, \(14x-7=0\Leftrightarrow x=\dfrac{1}{2}\)