Đáp án:
a, Ta có :
$\sqrt{1- 4x + 4x^2}$ = 5
<=> $\sqrt{(1-2x)^2}$ = 5
<=> $|1 - 2x| = 5$
<=> \(\left[ \begin{array}{l}1-2x=5\\1-2x = -5\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-2\\x=3\end{array} \right.\)
b, Ta có :
$\sqrt{x^2 + 6x + 9 }$ = 3x - 1
<=> $\sqrt{(x+3)^2}$ = 3x - 1
<=> $|x + 3| = 3x - 1$
<=> \(\left[ \begin{array}{l}x+3=3x-1\\x+3=1-3x\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=2\\x=-1/2\end{array} \right.\)
c, Ta có :
$\sqrt{x^2 + 4x + 4} - 2 = x$
<=>$\sqrt{(x+2)^2}$ = x+2
<=> | x + 2| = x + 2
<=> \(\left[ \begin{array}{l}x+2=x+2\\x+2=-x-2\end{array} \right.\)
<=> \(\left[ \begin{array}{l}∀x\\x=-2\end{array} \right.\)
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