Cho \({x_1}\) và \({x_2}\) là hai giá trị thỏa mãn \(x\left( {5 - 10x} \right) - 3\left( {10x - 5} \right) = 0\) . Khi đó \({x_1} + {x_2}\) bằng

\(m\left( {x + y + 1} \right)\).

\(m\left( {x + y + m} \right)\).

\(m\left( {x + y} \right)\).

\(m\left( {x + y - 1} \right)\)

\(4{x^3}{y^2} - 8{x^2}{y^3} = 4{x^2}y\left( {xy - 2{y^2}} \right)\).

\(4{x^3}{y^2} - 8{x^2}{y^3} = 4{x^2}{y^2}\left( {x - y} \right)\).

\(4{x^3}{y^2} - 8{x^2}{y^3} = 4{x^2}{y^2}\left( {x - 2y} \right)\).

\(4{x^3}{y^2} - 8{x^2}{y^3} = 4{x^2}{y^3}\left( {x - 2y} \right)\).

\({\left( {x - 2} \right)^2} - {\left( {2 - x} \right)^3} = {\left( {x - 2} \right)^2}\left( {x - 1} \right)\).

\({\left( {x - 2} \right)^2} - \left( {2 - x} \right) = \left( {x - 2} \right)\left( {x - 1} \right)\).

\({\left( {x - 2} \right)^3} - {\left( {2 - x} \right)^2} = {\left( {x - 2} \right)^2}\left( {3 - x} \right)\).

\({\left( {x - 2} \right)^2} + x - 2 = \left( {x - 2} \right)\left( {x - 1} \right)\).

\(5x\left( {x - y} \right) - \left( {y - x} \right) = \left( {x - y} \right)\left( {5x + 1} \right)\).

\(5x\left( {x - y} \right) - \left( {y - x} \right) = 5x\left( {x - y} \right)\).

\(5x\left( {x - y} \right) - \left( {y - x} \right) = \left( {x - y} \right)\left( {5x - 1} \right)\).  

\(5x\left( {x - y} \right) - \left( {y - x} \right) = \left( {x + y} \right)\left( {5x - 1} \right)\).

\(2a + b\).

\(1 + b\).

\({a^2} + ab\).

\(a + b\).

\(x + 2\).

\(3\left( {x - 2} \right)\).

\({\left( {x - 2} \right)^2}\).

\({\left( {x + 2} \right)^2}\).

\(x = 3;\,x =  - \dfrac{1}{2}\).

\(x =  - 3;\,x =  - \dfrac{1}{2}\).

\(x = 3;\,x = \dfrac{1}{2}\).

\(x =  - 3;\,x = \dfrac{1}{2}\).