Toán Lớp 9: $\sqrt{x+5}$ +$\sqrt{x+2}$ +2x-1=0
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Toán Lớp 9: $\sqrt{x+5}$ +$\sqrt{x+2}$ +2x-1=0
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\quad \sqrt{x+5} + \sqrt{x+2} + 2x – 1 =0\qquad (x\geqslant -2)\\
\Leftrightarrow \left(\sqrt{x+5} -2\right) + \left(\sqrt{x+2} -1\right) + 2x + 2=0\\
\Leftrightarrow \dfrac{\left(\sqrt{x+5} -2\right) \left(\sqrt{x+5} +2\right) }{\sqrt{x+5}+2} + \dfrac{ \left(\sqrt{x+2} -1\right) \left(\sqrt{x+2} +1\right)}{\sqrt{x+2} +1} + 2(x+1)= 0\\
\Leftrightarrow \dfrac{x+1}{\sqrt{x+5}+2} + \dfrac{x+1}{\sqrt{x+2} +1} + 2(x+1)= 0\\
\Leftrightarrow (x+1)\left(\dfrac{1}{\sqrt{x+5}+1} + \dfrac{1}{\sqrt{x+1}+1} +2\right) =0\\
\Leftrightarrow \left[\begin{array}{l}x = -1\qquad \text{(nhận)}\\\dfrac{1}{\sqrt{x+5}+1} + \dfrac{1}{\sqrt{x+1}+1} +2 =0\quad (vn)\end{array}\right.\\
\text{Vậy}\ S = \{-1\}
\end{array}\)