cho P=1-căn(x) tìm các giá trị x thỏa mãn: P.(căn(x)+1)=-(căn(3)+3).căn(x)+2căn(3)+3 điều kiện: x>0,x khác 4 24/09/2021 Bởi Alaia cho P=1-căn(x) tìm các giá trị x thỏa mãn: P.(căn(x)+1)=-(căn(3)+3).căn(x)+2căn(3)+3 điều kiện: x>0,x khác 4
Đáp án: $\begin{array}{l}Dkxd:x \ge 0\\P.\left( {\sqrt x + 1} \right) = – \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\ \Rightarrow \left( {1 – \sqrt x } \right).\left( {\sqrt x + 1} \right)\\ = – \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\ \Rightarrow 1 – x = – \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\ \Rightarrow x – \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 2 = 0\\ \Rightarrow x – 2\sqrt x – \left( {\sqrt 3 + 1} \right).\sqrt x + 2\left( {\sqrt 3 + 1} \right) = 0\\ \Rightarrow \left( {\sqrt x – 2} \right).\left( {\sqrt x – \sqrt 3 – 1} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}\sqrt x = 2\\\sqrt x = \sqrt 3 – 1\end{array} \right.\\ \Rightarrow \left[ \begin{array}{l}x = 4\left( {tm} \right)\\x = {\left( {\sqrt 3 – 1} \right)^2} = 4 – 2\sqrt 3 \left( {tm} \right)\end{array} \right.\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0\\
P.\left( {\sqrt x + 1} \right) = – \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\
\Rightarrow \left( {1 – \sqrt x } \right).\left( {\sqrt x + 1} \right)\\
= – \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\
\Rightarrow 1 – x = – \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 3\\
\Rightarrow x – \left( {\sqrt 3 + 3} \right).\sqrt x + 2\sqrt 3 + 2 = 0\\
\Rightarrow x – 2\sqrt x – \left( {\sqrt 3 + 1} \right).\sqrt x + 2\left( {\sqrt 3 + 1} \right) = 0\\
\Rightarrow \left( {\sqrt x – 2} \right).\left( {\sqrt x – \sqrt 3 – 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 2\\
\sqrt x = \sqrt 3 – 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4\left( {tm} \right)\\
x = {\left( {\sqrt 3 – 1} \right)^2} = 4 – 2\sqrt 3 \left( {tm} \right)
\end{array} \right.
\end{array}$