giải hpt 2x^2 - 4x + căn(y+1) =0 3x^2 - 6x -2căn(y+1) =-7

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

$\begin{array}{l} \left\{ \begin{array}{l} 2{x^2} - 4x + \sqrt {y + 1} = 0\\ 3{x^2} - 6x - 2\sqrt {y + 1} = - 7 \end{array} \right.\\ DK:\,\,\,y \ge - 1.\\ HPt \Leftrightarrow \left\{ \begin{array}{l} 2\left( {{x^2} - 2x} \right) + \sqrt {y + 1} = 0\\ 3\left( {{x^2} - 2x} \right) - 2\sqrt {y + 1} = - 7 \end{array} \right.\\ Dat\,\,\left\{ \begin{array}{l} a = {x^2} - 2x\\ b = \sqrt {y + 1} \,\,\,\left( {b \ge 0} \right) \end{array} \right.\\ \Rightarrow hpt \Leftrightarrow \left\{ \begin{array}{l} 2a + b = 0\\ 3a - 2b = - 7 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = - 1\\ b = 2\,\,\,\,\left( {tm} \right) \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 2x = - 1\\ \sqrt {y + 1} = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 2x + 1 = 0\\ y + 1 = 4 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {\left( {x - 1} \right)^2} = 0\\ y = 3\,\,\left( {tm} \right) \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ y = 3 \end{array} \right..\\ Vay\,\,hpt\,\,\,co\,\,nghiem\,\,\,\left( {x;\,\,y} \right) = \left( {1;\,\,3} \right). \end{array}$