Giải hộ mình với nhé 1) 4x^2+6x+7+√(2x^2+3x+9)=15 2) √(x+3) + √(6-x) – √(x+3)(6-x)=3 06/09/2021 Bởi Camila Giải hộ mình với nhé 1) 4x^2+6x+7+√(2x^2+3x+9)=15 2) √(x+3) + √(6-x) – √(x+3)(6-x)=3
Đáp án: $\begin{array}{l}1)\\4{x^2} + 6x + 7 + \sqrt {2{x^2} + 3x + 9} = 15\\ \Rightarrow 2\left( {2{x^2} + 3x + 9} \right) – 11 + \sqrt {2{x^2} + 3x + 9} – 15 = 0\\ \Rightarrow 2.\left( {2{x^2} + 3x + 9} \right) + \sqrt {2{x^2} + 3x + 9} – 26 = 0\\Đặt:\sqrt {2{x^2} + 3x + 9} = t\left( {t \ge 0} \right)\\ \Rightarrow 2{t^2} + t – 26 = 0\\ \Rightarrow t = \dfrac{{\sqrt {209} – 1}}{4}\left( {do:t \ge 0} \right)\\ \Rightarrow 2{x^2} + 3x + 9 = {t^2} = 11,318\\ \Rightarrow 2{x^2} + 3x – 2,318 = 0\\ \Rightarrow \left[ \begin{array}{l}x = – 2,062\\x = 0,562\end{array} \right.\\2)\sqrt {x + 3} + \sqrt {6 – x} – \sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = 3\\\left( {Dkxd: – 3 \le x \le 6} \right)\\Dat:\sqrt {x + 3} + \sqrt {6 – x} = t\\ \Rightarrow x + 3 + 2\sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} + 6 – x = {t^2}\\ \Rightarrow 2\sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = {t^2} – 9\left( {dk:{t^2} \ge 9} \right)\\ \Rightarrow \sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = \dfrac{{{t^2} – 9}}{2}\\ \Rightarrow PT:t – \dfrac{{{t^2} – 9}}{2} = 3\\ \Rightarrow {t^2} – 2t – 3 = 0\\ \Rightarrow \left( {t – 3} \right)\left( {t + 1} \right) = 0\\ \Rightarrow t = 3\\ \Rightarrow \sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = 0\\ \Rightarrow \left[ \begin{array}{l}x = – 3\\x = 6\end{array} \right.\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
1)\\
4{x^2} + 6x + 7 + \sqrt {2{x^2} + 3x + 9} = 15\\
\Rightarrow 2\left( {2{x^2} + 3x + 9} \right) – 11 + \sqrt {2{x^2} + 3x + 9} – 15 = 0\\
\Rightarrow 2.\left( {2{x^2} + 3x + 9} \right) + \sqrt {2{x^2} + 3x + 9} – 26 = 0\\
Đặt:\sqrt {2{x^2} + 3x + 9} = t\left( {t \ge 0} \right)\\
\Rightarrow 2{t^2} + t – 26 = 0\\
\Rightarrow t = \dfrac{{\sqrt {209} – 1}}{4}\left( {do:t \ge 0} \right)\\
\Rightarrow 2{x^2} + 3x + 9 = {t^2} = 11,318\\
\Rightarrow 2{x^2} + 3x – 2,318 = 0\\
\Rightarrow \left[ \begin{array}{l}
x = – 2,062\\
x = 0,562
\end{array} \right.\\
2)\sqrt {x + 3} + \sqrt {6 – x} – \sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = 3\\
\left( {Dkxd: – 3 \le x \le 6} \right)\\
Dat:\sqrt {x + 3} + \sqrt {6 – x} = t\\
\Rightarrow x + 3 + 2\sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} + 6 – x = {t^2}\\
\Rightarrow 2\sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = {t^2} – 9\left( {dk:{t^2} \ge 9} \right)\\
\Rightarrow \sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = \dfrac{{{t^2} – 9}}{2}\\
\Rightarrow PT:t – \dfrac{{{t^2} – 9}}{2} = 3\\
\Rightarrow {t^2} – 2t – 3 = 0\\
\Rightarrow \left( {t – 3} \right)\left( {t + 1} \right) = 0\\
\Rightarrow t = 3\\
\Rightarrow \sqrt {\left( {x + 3} \right)\left( {6 – x} \right)} = 0\\
\Rightarrow \left[ \begin{array}{l}
x = – 3\\
x = 6
\end{array} \right.
\end{array}$