Giải thích các bước giải:
a) `sqrt(2x-1)+1=2x`
`=>sqrt(2x-1)-(2x-1)=0`
`=>sqrt(2x-1).(1-sqrt(2x-1))=0`
`=>`\(\left[ \begin{array}{l}2x-1=0\\2x-1=1\end{array} \right.\) `=>`\(\left[ \begin{array}{l}2x=0\\2x=1\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac12\end{array} \right.\)
Vậy `x in {0;1/2}.`
b) `sqrt(3x-2)+4 =6x`
`=>sqrt(3x-2)-(6x-4)=0`
`=>sqrt(3x-2)-2(3x-2)=0`
`=>sqrt(3x-2).(1-2sqrt(3x-2))=0`
`=>`\(\left[ \begin{array}{l}3x-2=0\\3x-2=1\end{array} \right.\) `=>`\(\left[ \begin{array}{l}3x=2\\3x=3\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\dfrac23\\x=1\end{array} \right.\)
Vậy `x in {1;2/3}.`
