Trắc nghiệm Xác định phương trình và hệ phương trình bậc nhất hai ẩn Toán 9 có đáp án

$2{x^2} + 2 = 0$

$3y - 1 = 5y\left( {y - 2} \right)$

$2x + \dfrac{y}{2} - 1 = 0$

$3\sqrt x  + {y^2} = 0$

$\left\{ \begin{array}{l}2a + 5b = \dfrac{5}{6}\\3a - 4b =  - \dfrac{3}{5}\end{array} \right.$

$\left\{ \begin{array}{l}2a + 5b = \dfrac{6}{5}\\3a - 4b =  - \dfrac{5}{3}\end{array} \right.$

$\left\{ \begin{array}{l}2a - 5b = \dfrac{5}{6}\\3a + 4b =  - \dfrac{3}{5}\end{array} \right.$

$\left\{ \begin{array}{l}-2a - 5b = \dfrac{5}{6}\\3a - 4b =  - \dfrac{3}{5}\end{array} \right.$

$\left\{ \begin{array}{l}15a - 7b = 9\\ - 4a + 9b = 5\end{array} \right.$

$\left\{ \begin{array}{l}15a - 7b = 9\\4a + 9b = 5\end{array} \right.$

$\left\{ \begin{array}{l}15a - 7b =  - 9\\4a + 9b = \dfrac{1}{5}\end{array} \right.$

$\left\{ \begin{array}{l} - 15a + 7b = 9\\4a - 9b = 5\end{array} \right.$

$4x + 0y - 6 = 0$

$\sqrt x  + x - 1 = 0$

${x^2} + \dfrac{y}{2} = 0$

${x^3} + 1 = 0$

$\left\{ \begin{array}{l}\dfrac{1}{2}a + \dfrac{1}{6}b = 3\\\dfrac{1}{4}a - \dfrac{1}{9}b = 1\end{array} \right.$

$\left\{ \begin{array}{l}2a + 6b = 3\\4a - 9b = 1\end{array} \right.$

$\left\{ \begin{array}{l}2b + 6a = 3\\4b - 9a = 1\end{array} \right.$

$\left\{ \begin{array}{l}\dfrac{2}{3}a + 6b = 3\\4a - 9b = 1\end{array} \right.$

\(a \ne 0\) và \(b \ne 0\).

\(a \ne 0\) hoặc \(b \ne 0\).

\(a = 0;b = 0\) và \(c \ne 0\)

\(c \ne 0\).

\(2x + 3y = 4\).

\(3x + 0y = 5\).

\(0x + 4y = 8\).

\(0x + 0y =  - 3\).

\(a = 2;b =  - 1;c = 3\).

\(a = 2;b = 1;c = 3\).

\(a = 2;b =  - 1;c =  - 3\).

\(a = 2;b = 1;c =  - 3\).

\(\left\{ \begin{array}{l}a{x^2} + b{y^2} = c\\a'x + b'y = c'\end{array} \right.\).

\(\left\{ \begin{array}{l}ax + by = c\\a'{x^2} + b'{y^2} = c'\end{array} \right.\).

\(\left\{ \begin{array}{l}a{x^2} + by = c\\a'x + b'{y^2} = c'\end{array} \right.\).

\(\left\{ \begin{array}{l}ax + by = c\\a'x + b'y = c'\end{array} \right.\).

\(\left\{ \begin{array}{l}2x + y = 3\\x - 2y = 5\end{array} \right.\)

\(\left\{ \begin{array}{l}x + 2y = 4\\{x^2} + y = 1\end{array} \right.\)

\(\left\{ \begin{array}{l}2x + 0y = 3\\x + y = 2\end{array} \right.\)

\(\left\{ \begin{array}{l}x + 2y = 3\\0x - y = 1\end{array} \right.\)

\(\left\{ \begin{array}{l}x + y = 36\\2x + 4y = 100\end{array} \right.\).

\(\left\{ \begin{array}{l}x + y = 36\\4x + 2y = 100\end{array} \right.\).

\(\left\{ \begin{array}{l}2x + 4y = 36\\x + y = 100\end{array} \right.\).

\(\left\{ \begin{array}{l}4x + 2y = 36\\x + y = 100\end{array} \right.\).

\(2x + 3y = 5\).

\(0x + 2y = 8\).

\(2x - 0y = 5\).

\(0x - 0y = 6\).

\(\left\{ \begin{array}{l}x + 12y = 3\\x - 15y = 12\end{array} \right.\).

\(\left\{ \begin{array}{l}x - 12y = 3\\ - x + 15y = 12\end{array} \right.\).

\(\left\{ \begin{array}{l}x - 12y = 3\\x + 15y = 12\end{array} \right.\).

\(\left\{ \begin{array}{l}x + 12y = 3\\ - x + 15y = 12\end{array} \right.\).

\(\left\{ \begin{array}{l}2x - 3y = 5\\x + 3y =  - 11\end{array} \right.\).

\(\left\{ \begin{array}{l}2x - 3y = 5\\3x =  - 6\end{array} \right.\).

\(\left\{ \begin{array}{l}9y =  - 27\\x + 3y =  - 11\end{array} \right.\).

\(\left\{ \begin{array}{l}{x^2} + {y^2} = 121\\x + 2y =  - 11\end{array} \right.\).

\(a = 12,b = 5,c = 4\).

\(a = 12,b =  - 5,c =  - 4\).

\(a =  - 12,b =  - 5,c =  - 4\).

\(a = 12,b =  - 5,c = 4\).

\(xy + x = 3\).

\(x + y = xy\).

\(2x - y = 0\).

\({x^2} + {y^2} = 5\).

\(2x + 3{y^2} = 0\).

\(xy - x = 1\).

\(3x + 2{y^3} = 1\).

\(3x - y = 5\).