Đáp án:
a, x = 3
b, \(\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\)
c, x = 2
Giải thích các bước giải:
a, ĐK : x ≥ 2
pt ⇔ x² - 6x + 9 + x + 2 - 2$\sqrt{x+2}$ + 1 = 0 ⇔ ( x - 3 )² + ( $\sqrt{x+2}$ - 1 )² = 0
⇔ \(\left[ \begin{array}{l}x=3\\\sqrt{x+2}=1\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=3\\x-2=1\end{array} \right.\) ⇔ x = 3 ( tm ĐK )
b, ĐK : x ≥ -3
đặt $\sqrt{x+3}$ = a ( a ≥ 0 )
$\sqrt{x+5 }$ = b ( b ≥ 0 )
pt ⇔ ab = 3a + 2b - 6 ⇔ ( 3 - b ) ( a - 2 ) = 0
⇔ \(\left[ \begin{array}{l}\sqrt{x+3}=2\\\sqrt{x+5 }=3\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x+3=4\\x+5=9\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=4\\x=1\end{array} \right.\)
c, ĐK : x ≥ $\dfrac{3}{2}$
pt ⇔ x² - 4x + 4 + 2x - 3 - 2$\sqrt{2x-3}$ + 1 = 0
⇔ ( x - 2 )² + ( $\sqrt{2x-3}$ -1 )² =0
⇔ \(\left[ \begin{array}{l}x=2\\\sqrt{2x-3} =1\end{array} \right.\) ⇔ \(\left[ \begin{array}{l}x=2\\2x-3=1\end{array} \right.\) ⇔ x = 2
