Đáp án:
Giải thích các bước giải :
`a.2x(x-3) - 3(x-3) = 0`
`⇔(x-3)(2x-3) = 0`
⇔\(\left[ \begin{array}{l}x-3=0\\2x-3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=3/2\end{array} \right.\)
Tập nghiệm `S =3, 3/2`
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`b.x²(x-1)+4(1-x) =0 `
`⇔ x²(x-1) - 4(x-1) =0 `
`⇔ (x-1)(x²-4) = 0`
\(\left[ \begin{array}{l}x-1=0\\x^2-4=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\)
Vậy ......
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`c.2x(x-5) = (x-5)²`
`⇔ 2x(x-5) - (x-5)² =0`
`⇔ (x-5)[2x-(x-5)] = 0`
`⇔ (x-5)(x+5) =0`
\(\left[ \begin{array}{l}x-5=0\\x+5=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=5\\x=-5\end{array} \right.\)
Vậy ....
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`d. (2x-1)² = (4-3x)²`
`⇔ 4x² - 4x + 1 = 16 - 24x + 9x²`
`⇔ 4x² -4x+1 - 16+24x-9x² = 0`
`⇔ -5x² + 20x - 15 = 0`
`⇔x² - 4x+3 = 0`
`⇔ x² -x -3x+3 = 0`
`⇔ x(x-1) - 3(x-1) = 0`
`⇔ (x-1)(x-3) = 0`
\(\left[ \begin{array}{l}x-1=0\\x-3=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=1\\x=3\end{array} \right.\)
Vậy .....
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`e. 2x ( 3 - 4x ) - 5x²( 4x - 3 ) = 0`
`⇔ 2x ( 3 - 4x ) + 5x²( 3-4x) = 0`
`⇔ (3-4x)(2x+ 5x² ) =0 `
`⇔ x(3-4x) (2+5x) =0 `
`x=0`
\(\left[ \begin{array}{l}3-4x=0\\2-5x=0\end{array} \right.\)
\(\left[ \begin{array}{l}x=3/4\\x=-2/5\end{array} \right.\)
Chúc bn học tốt !!!!
