Toán giải hpt 2x^2 – 4x + căn(y+1) =0 3x^2 – 6x -2căn(y+1) =-7 04/10/2021 By Jade giải hpt 2x^2 – 4x + căn(y+1) =0 3x^2 – 6x -2căn(y+1) =-7
\[\begin{array}{l} \left\{ \begin{array}{l} 2{x^2} – 4x + \sqrt {y + 1} = 0\\ 3{x^2} – 6x – 2\sqrt {y + 1} = – 7 \end{array} \right.\\ DK:\,\,\,y \ge – 1.\\ HPt \Leftrightarrow \left\{ \begin{array}{l} 2\left( {{x^2} – 2x} \right) + \sqrt {y + 1} = 0\\ 3\left( {{x^2} – 2x} \right) – 2\sqrt {y + 1} = – 7 \end{array} \right.\\ Dat\,\,\left\{ \begin{array}{l} a = {x^2} – 2x\\ b = \sqrt {y + 1} \,\,\,\left( {b \ge 0} \right) \end{array} \right.\\ \Rightarrow hpt \Leftrightarrow \left\{ \begin{array}{l} 2a + b = 0\\ 3a – 2b = – 7 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = – 1\\ b = 2\,\,\,\,\left( {tm} \right) \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} – 2x = – 1\\ \sqrt {y + 1} = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^2} – 2x + 1 = 0\\ y + 1 = 4 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {\left( {x – 1} \right)^2} = 0\\ y = 3\,\,\left( {tm} \right) \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ y = 3 \end{array} \right..\\ Vay\,\,hpt\,\,\,co\,\,nghiem\,\,\,\left( {x;\,\,y} \right) = \left( {1;\,\,3} \right). \end{array}\] Trả lời
Đáp án: Giải thích các bước giải: \(\left\{ \begin{array}{l}2{x^2} – 4x + \sqrt {y + 1} = 0\\3{x^2} – 6x – 2\sqrt {y + 1} = – 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2\left( {{x^2} – 2x} \right) + \sqrt {y + 1} = 0\\3\left( {{x^2} – 2x} \right) + 2.\sqrt {y + 1} = – 7\end{array} \right.\) Sau đó đặt ẩn em nhé Trả lời
\[\begin{array}{l}
\left\{ \begin{array}{l}
2{x^2} – 4x + \sqrt {y + 1} = 0\\
3{x^2} – 6x – 2\sqrt {y + 1} = – 7
\end{array} \right.\\
DK:\,\,\,y \ge – 1.\\
HPt \Leftrightarrow \left\{ \begin{array}{l}
2\left( {{x^2} – 2x} \right) + \sqrt {y + 1} = 0\\
3\left( {{x^2} – 2x} \right) – 2\sqrt {y + 1} = – 7
\end{array} \right.\\
Dat\,\,\left\{ \begin{array}{l}
a = {x^2} – 2x\\
b = \sqrt {y + 1} \,\,\,\left( {b \ge 0} \right)
\end{array} \right.\\
\Rightarrow hpt \Leftrightarrow \left\{ \begin{array}{l}
2a + b = 0\\
3a – 2b = – 7
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = – 1\\
b = 2\,\,\,\,\left( {tm} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} – 2x = – 1\\
\sqrt {y + 1} = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{x^2} – 2x + 1 = 0\\
y + 1 = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – 1} \right)^2} = 0\\
y = 3\,\,\left( {tm} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 1\\
y = 3
\end{array} \right..\\
Vay\,\,hpt\,\,\,co\,\,nghiem\,\,\,\left( {x;\,\,y} \right) = \left( {1;\,\,3} \right).
\end{array}\]
Đáp án:
Giải thích các bước giải:
\(\left\{ \begin{array}{l}2{x^2} – 4x + \sqrt {y + 1} = 0\\3{x^2} – 6x – 2\sqrt {y + 1} = – 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2\left( {{x^2} – 2x} \right) + \sqrt {y + 1} = 0\\3\left( {{x^2} – 2x} \right) + 2.\sqrt {y + 1} = – 7\end{array} \right.\)
Sau đó đặt ẩn em nhé