Toán Tìm x, biết: a)4x^2-9=x(2x-3) b)(x-1)^2+x(4-x)=0 c)x(x+2)-3x-6=0 02/09/2021 By Peyton Tìm x, biết: a)4x^2-9=x(2x-3) b)(x-1)^2+x(4-x)=0 c)x(x+2)-3x-6=0
a) $4x^{2}$-9=x(2x-3) ⇔ $(2x)^{2}$-$3^{2}$=x(2x-3) ⇔(2x-3)(2x+3)=x(2x-3) ⇔ (2x-3)(2x+3-x)=0 ⇔ (2x-3)(x+3)=0 2x-3=0 hoặc x+3=0 2x=3 hoặc x=-3 x=$\frac{3}{2}$ hoặc x=-3 b) $(x-1)^{2}$+x(4-x)=0 ⇔ $x^{2}$-2x+1+4x-$x^{2}$=0 ⇔ 2x+1=0 ⇔x=$\frac{-1}{2}$ c) x(x+2)-3x-6=0 ⇔ x(x+2)+3(x+2)=0 ⇔ (x+2)(x+3)=0 x+2=0 hoặc x+3=0 x=-2hoặc x=-3 Trả lời
Đáp án: $\begin{array}{l}a)4{x^2} – 9 = x\left( {2x – 3} \right)\\ \Rightarrow {\left( {2x} \right)^2} – {3^2} = x\left( {2x – 3} \right)\\ \Rightarrow \left( {2x – 3} \right)\left( {2x + 3} \right) – x\left( {2x – 3} \right) = 0\\ \Rightarrow \left( {2x – 3} \right)\left( {2x + 3 – x} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}2x – 3 = 0\\x + 3 = 0\end{array} \right. \Rightarrow \left[ \begin{array}{l}x = \frac{3}{2}\\x = – 3\end{array} \right.\\b){\left( {x – 1} \right)^2} + x\left( {4 – x} \right) = 0\\ \Rightarrow {x^2} – 2x + 1 + 4x – {x^2} = 0\\ \Rightarrow 2x + 1 = 0\\ \Rightarrow x = \frac{{ – 1}}{2}\\c)x\left( {x + 2} \right) – 3x – 6 = 0\\ \Rightarrow x\left( {x + 2} \right) – 3\left( {x + 2} \right) = 0\\ \Rightarrow \left( {x + 2} \right)\left( {x – 3} \right) = 0\\ \Rightarrow \left[ \begin{array}{l}x = – 2\\x = 3\end{array} \right.\end{array}$ Trả lời
a) $4x^{2}$-9=x(2x-3)
⇔ $(2x)^{2}$-$3^{2}$=x(2x-3)
⇔(2x-3)(2x+3)=x(2x-3)
⇔ (2x-3)(2x+3-x)=0
⇔ (2x-3)(x+3)=0
2x-3=0 hoặc x+3=0
2x=3 hoặc x=-3
x=$\frac{3}{2}$ hoặc x=-3
b) $(x-1)^{2}$+x(4-x)=0
⇔ $x^{2}$-2x+1+4x-$x^{2}$=0
⇔ 2x+1=0
⇔x=$\frac{-1}{2}$
c) x(x+2)-3x-6=0
⇔ x(x+2)+3(x+2)=0
⇔ (x+2)(x+3)=0
x+2=0 hoặc x+3=0
x=-2hoặc x=-3
Đáp án:
$\begin{array}{l}
a)4{x^2} – 9 = x\left( {2x – 3} \right)\\
\Rightarrow {\left( {2x} \right)^2} – {3^2} = x\left( {2x – 3} \right)\\
\Rightarrow \left( {2x – 3} \right)\left( {2x + 3} \right) – x\left( {2x – 3} \right) = 0\\
\Rightarrow \left( {2x – 3} \right)\left( {2x + 3 – x} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
2x – 3 = 0\\
x + 3 = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \frac{3}{2}\\
x = – 3
\end{array} \right.\\
b){\left( {x – 1} \right)^2} + x\left( {4 – x} \right) = 0\\
\Rightarrow {x^2} – 2x + 1 + 4x – {x^2} = 0\\
\Rightarrow 2x + 1 = 0\\
\Rightarrow x = \frac{{ – 1}}{2}\\
c)x\left( {x + 2} \right) – 3x – 6 = 0\\
\Rightarrow x\left( {x + 2} \right) – 3\left( {x + 2} \right) = 0\\
\Rightarrow \left( {x + 2} \right)\left( {x – 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = – 2\\
x = 3
\end{array} \right.
\end{array}$