Question

Giai pt

Answer 1

1: ĐKXĐ: x>=3/2
\(\sqrt{2x-3}=1\)

=>\(2x-3=1^2=1\)

=>2x=4

=>x=2(nhận)

2: ĐKXĐ: x∈R

\(\sqrt{4x^2-12x+9}=x\)

=>\(\sqrt{\left(2x-3\right)^2}=x\)

=>|2x-3|=x

=>\(\begin{cases}x\ge0\\ \left.\left(2x-3\right)^2=x^2\right.\end{cases}\Rightarrow\begin{cases}x\ge0\\ \left(2x-3-x\right)\left(2x-3+x\right)=0\end{cases}\)

=>\(\begin{cases}x\ge0\\ \left(x-3\right)\left(3x-3\right)=0\end{cases}\Rightarrow x\in\left\lbrace3;1\right\rbrace\)