Tìm X x^2(x-3)+12-4x=0 2(x+5)-x^2-5x=0 9(x-1)^2-4(x+3)^2=0

Đáp án:

Giải thích các bước giải: $\begin{array}{l} + )\,{x^2}\left( {x - 3} \right) + 12 - 4x = 0\\ \Leftrightarrow {x^2}\left( {x - 3} \right) - 4\left( {x - 3} \right) = 0\\ \Leftrightarrow \left( {x - 3} \right)\left( {{x^2} - 4} \right) = 0\\ \Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right)\left( {x - 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x - 3 = 0\\x + 2 = 0\\x - 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\\x = - 2\\x = 2\end{array} \right.\\ + )\,2\left( {x + 5} \right) - {x^2} - 5x = 0\\ \Leftrightarrow 2\left( {x + 5} \right) - x\left( {x + 5} \right) = 0\\ \Leftrightarrow \left( {x + 5} \right)\left( {2 - x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x + 5 = 0\\2 - x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 5\\x = 2\end{array} \right.\\ + )\,9{\left( {x - 1} \right)^2} - 4{\left( {x + 3} \right)^2} = 0\\ \Leftrightarrow 9\left( {{x^2} - 2x + 1} \right) - 4\left( {{x^2} + 6x + 9} \right) = 0\\ \Leftrightarrow 9{x^2} - 18x + 9 - 4{x^2} - 24x - 36 = 0\\ \Leftrightarrow 5{x^2} - 42x - 27 = 0\\ \Leftrightarrow \left( {x - 9} \right)\left( {5x + 3} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}x - 9 = 0\\5x + 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 9\\x = \frac{{ - 3}}{5}\end{array} \right.\end{array}$

1) $\Rightarrow x^2(x-3)-4(x-3)=0$ $\Rightarrow (x-3)(x^2-4)=0$ $\Rightarrow(x-3)(x+2)(x-2)=0$ $\Rightarrow \left[ \begin{array}{l} x=3 \\x=-2\\x=2 \end{array} \right .$ 2) $2(x+5)-x(x+5)=0$ $(x+5)(2-x)=0$ $\left[ \begin{array}{l} x=-5 \\ x=2\end{array} \right .$ 3) $[3(x-1)+2(x+3)][3(x-1)-2(x+3)]=0$ $(5x+3)(x-9)=0$ $\Rightarrow \left[ \begin{array}{l} x=\dfrac{-3}{5} \\x=9\end{array} \right .$