Question

Giải rõ ra nhé !! Tớ sẽ tick sau

Answer 1

a: \(\sqrt{3-2x}=x\)(ĐKXĐ: \(x< =\dfrac{3}{2}\))

=>\(\left\{{}\begin{matrix}x>=0\\3-2x=x^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0< =x< =\dfrac{3}{2}\\x^2+2x-3=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}0< =x< =1,5\\\left(x+3\right)\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow x=1\)

=>Sai

b: ĐKXĐ: x>=-11/7

\(\sqrt{7x+11}+x+1=0\)

=>\(\sqrt{7x+11}=-x-1\)

=>\(\left\{{}\begin{matrix}-x-1>=0\\\left(-x-1\right)^2=7x+11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =-1\\x^2+2x+1-7x-11=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{11}{7}< =x< =-1\\x^2-5x-10=0\end{matrix}\right.\Leftrightarrow x=\dfrac{5-\sqrt{65}}{2}\)

=>Sai

c: Đúng

d: Đúng