Question

giải phuơng trình

\(a.\sqrt{4x^2}=3\)

\(b.\sqrt{4x^2}=x-1\)

\(c.\sqrt{x-1}=2x-1\)

Answer 1

\(a,\sqrt{\left(4x\right)^2}=3^2\)

=> \(\left|4x\right|=9\)

=> \(\left[{}\begin{matrix}4x=9\\4x=-9\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{9}{4}\\x=-\frac{9}{4}\end{matrix}\right.\)

\(b,\sqrt{4x^2}=x-1\)

=> \(\sqrt{\left(4x^2\right)^2}=x-1\)

=> \(\left|4x^2\right|=x-1\)

\(\)=> \(\left[{}\begin{matrix}4x^2=x-1\\4x^2=x+1\end{matrix}\right.\)

=> :))

c, :))

Answer 2

\(a.\)\(\left(\sqrt{4x^2}\right)^2=3^2\Leftrightarrow4x^2=9\Leftrightarrow x^2=\frac{9}{4}\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy ...............

\(b.\left(\sqrt{4x^2}\right)^2=\left(x-1\right)^2\Leftrightarrow4x^2=x^2-2x+1\Leftrightarrow3x^2+2x-1=0\Leftrightarrow\left(x-\frac{1}{3}\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-1\end{matrix}\right.\)

Vậy ..............