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 19/09/2021 Bởi Emery 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
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) $[3(x-1)+2(x+3)][3(x-1)-2(x+3)]=0$ $(5x+3)(x-9)=0$ $\Rightarrow \left[ \begin{array}{l} x=\dfrac{-3}{5} \\x=9\end{array} \right .$ Bình luận
Đáp án: Giải thích các bước giải: \(\begin{array}{l} + )\,{x^2}\left( {x – 3} \right) + 12 – 4x = 0\\ \Leftrightarrow {x^2}\left( {x – 3} \right) – 4\left( {x – 3} \right) = 0\\ \Leftrightarrow \left( {x – 3} \right)\left( {{x^2} – 4} \right) = 0\\ \Leftrightarrow \left( {x – 3} \right)\left( {x + 2} \right)\left( {x – 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x – 3 = 0\\x + 2 = 0\\x – 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\\x = – 2\\x = 2\end{array} \right.\\ + )\,2\left( {x + 5} \right) – {x^2} – 5x = 0\\ \Leftrightarrow 2\left( {x + 5} \right) – x\left( {x + 5} \right) = 0\\ \Leftrightarrow \left( {x + 5} \right)\left( {2 – x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x + 5 = 0\\2 – x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = – 5\\x = 2\end{array} \right.\\ + )\,9{\left( {x – 1} \right)^2} – 4{\left( {x + 3} \right)^2} = 0\\ \Leftrightarrow 9\left( {{x^2} – 2x + 1} \right) – 4\left( {{x^2} + 6x + 9} \right) = 0\\ \Leftrightarrow 9{x^2} – 18x + 9 – 4{x^2} – 24x – 36 = 0\\ \Leftrightarrow 5{x^2} – 42x – 27 = 0\\ \Leftrightarrow \left( {x – 9} \right)\left( {5x + 3} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}x – 9 = 0\\5x + 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 9\\x = \frac{{ – 3}}{5}\end{array} \right.\end{array}\) Bình luận
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) $[3(x-1)+2(x+3)][3(x-1)-2(x+3)]=0$
$(5x+3)(x-9)=0$
$\Rightarrow \left[ \begin{array}{l} x=\dfrac{-3}{5} \\x=9\end{array} \right .$
Đáp án:
Giải thích các bước giải:
\(\begin{array}{l} + )\,{x^2}\left( {x – 3} \right) + 12 – 4x = 0\\ \Leftrightarrow {x^2}\left( {x – 3} \right) – 4\left( {x – 3} \right) = 0\\ \Leftrightarrow \left( {x – 3} \right)\left( {{x^2} – 4} \right) = 0\\ \Leftrightarrow \left( {x – 3} \right)\left( {x + 2} \right)\left( {x – 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x – 3 = 0\\x + 2 = 0\\x – 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 3\\x = – 2\\x = 2\end{array} \right.\\ + )\,2\left( {x + 5} \right) – {x^2} – 5x = 0\\ \Leftrightarrow 2\left( {x + 5} \right) – x\left( {x + 5} \right) = 0\\ \Leftrightarrow \left( {x + 5} \right)\left( {2 – x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x + 5 = 0\\2 – x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = – 5\\x = 2\end{array} \right.\\ + )\,9{\left( {x – 1} \right)^2} – 4{\left( {x + 3} \right)^2} = 0\\ \Leftrightarrow 9\left( {{x^2} – 2x + 1} \right) – 4\left( {{x^2} + 6x + 9} \right) = 0\\ \Leftrightarrow 9{x^2} – 18x + 9 – 4{x^2} – 24x – 36 = 0\\ \Leftrightarrow 5{x^2} – 42x – 27 = 0\\ \Leftrightarrow \left( {x – 9} \right)\left( {5x + 3} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}x – 9 = 0\\5x + 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 9\\x = \frac{{ – 3}}{5}\end{array} \right.\end{array}\)