Toán Lớp 10: Giải phương trình
$\sqrt{-x^2+9x+9}=\sqrt{x}+\sqrt{9-x}$
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 10: Giải phương trình
$\sqrt{-x^2+9x+9}=\sqrt{x}+\sqrt{9-x}$
Comments ( 1 )
\quad \sqrt{-x^2+9x + 9} = \sqrt x +\sqrt{9 – x}\\
ĐKXĐ: \begin{cases}-x^2 + 9x + 9 \geqslant 0\\x \geqslant 0\\9 – x \geqslant 0\end{cases}\Leftrightarrow 0\leqslant x \leqslant 9\\
\text{Đặt}\ \begin{cases}a = \sqrt x\\b = \sqrt{9-x}\end{cases}\qquad (a;b\geqslant 0)\\
\text{Phương trình trở thành:}\\
\quad\sqrt{ab + a^2 + b^2} = a + b\\
\Rightarrow ab + a^2 + b^2 = (a+b)^2\\
\Leftrightarrow ab = 0\\
\Leftrightarrow \sqrt{x(9-x)} = 0\\
\Leftrightarrow \left[\begin{array}{l}x = 0\\x = 9\end{array}\right.\quad \text{(nhận)}\\
\text{Vậy}\ S =\{0;9\}
\end{array}\)