Đáp án:
1.
a, Ta có
$\sqrt{x^2 - 4} = x - 2$ ` ( ĐKXĐ : x ≥ 2)`
<=> $\sqrt{x^2 - 4}^2= (x - 2)^2$
`<=> x^2 - 4 = (x - 2)(x - 2)`
` <=> (x - 2)(x + 2) - (x - 2)(x - 2) = 0`
` <=> (x - 2)(x + 2 - x + 2) = 0`
` <=> (x - 2).4 = 0`
` <=> x - 2 = 0`
` <=> x = 2`
b, Ta có
$\sqrt{x^2 + 1} = x + 3 $ `(ĐKXĐ : ∀ mọi x)`
`<=> x^2 + 1 = x^2 + 6x + 9`
` <=> 6x + 9 = 1`
` <=> 6x = -8`
` <=> x = -4/3`
c, Ta có
$\sqrt{5 - x^2} = x - 1$ `(ĐKXĐ : -√5 ≤ x ≤ √5)`
` <=> 5 - x^2 = x^2 - 2x + 1`
` <=> 5 - x^2 - x^2 + 2x - 1 = 0`
` <=> -2x^2 + 2x + 4 = 0`
` <=> -x^2 + x + 2 = 0`
` <=> -x^2 + 2x - x + 2 = 0`
` <=> -x.(x - 2) - (x - 2) = 0`
` <=> (x - 2)(1 + x) = 0`
<=> \(\left[ \begin{array}{l}x - 2 = 0\\1 + x = 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
Bài 2.
a, Ta có
$\sqrt{x^2 - 1} - \sqrt{x - 1} = 0$ (ĐKXĐ : x $\neq$ 0)
$\sqrt{x^2 - 1} = \sqrt{x - 1}$
` <=> x^2 - 1 = x - 1`
` <=> x^2 - 1 - (x - 1) = 0`
` <=> (x - 1)(x + 1) - (x - 1) = 0`
` <=> (x - 1)(x + 1 - 1) = 0`
` <=> (x - 1)x = 0`
Do x $\neq$ 0
` <=> x - 1 = 0`
` <=> x = 1`
b, Ta có
`$\sqrt{x + 4} - \sqrt{x^2 + 5x + 4} = 0$ `(ĐKXĐ : x ≥ - 1)`
` <=> x + 4 - x^2 - 5x - 4 = 0`
` <=> -x^2 - 4x = 0`
` <=> -x.(x + 4) = 0`
<=> \(\left[ \begin{array}{l}-x = 0\\x + 4 = 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=0\\x=-4\end{array} \right.\)
Giải thích các bước giải:
