Đáp án:
a, `ĐKXĐ : ∀x`
Ta có
`\sqrt{|x - 1|} = x + 2`
`<=> |x - 1| = (x + 2)^2`
`<=> |x - 1| = x^2 + 4x + 4`
<=> \(\left[ \begin{array}{l}x - 1 = x^2 + 4x + 4 \\1 - x = x^2 + 4x + 4\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x^2 + 3x + 5 = 0\\x^2 + 5x + 3 = 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}<L>\\x= ± \sqrt{13}/2 - 5/2 \end{array} \right.\)
Do `x ≥ -2`
`<=> x = \sqrt{13}/2 - 5/2`
b, ` ĐKXĐ : x ≥ 1/2`
Ta có :
`\sqrt{2x - 1} = |x|`
`<=> 2x - 1 = x^2`
`<=> x^2 - 2x + 1 = 0`
`<=> (x - 1)^2 = 0`
`<=> x - 1 =0`
`<=> x = 1`
c, `ĐKXĐ : ∀x`
Ta có
`\sqrt{x^2 + 6x + 9} = |2x - 1|`
`<=> \sqrt{(x + 3)^2} = |2x - 1|`
`<=> |x + 3| = |2x - 1|`
<=> \(\left[ \begin{array}{l}x + 3 = 2x - 1\\x + 3 = 1 - 2x\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x= 4\\x= -2/3\end{array} \right.\)
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