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`e)`
`x^2-2x+1=4`
`<=>(x-1)^2=2^2`
`<=>` \(\left[ \begin{array}{l}x-1=2\\x-1=-2\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
Vậy `S={-1;3}`
``
`g)`
`2x^2-3x+1=0`
`<=>2x^2-2x-x+1=0`
`<=>2x(x-1)-(x-1)=0`
`<=>(2x-1)(x-1)=0`
`<=>` \(\left[ \begin{array}{l}2x-1=0\\x-1=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=1\end{array} \right.\)
Vậy `S={1/2;1}`
``
`h)`
`x^2-5x+6=0`
`<=>x^2-2x-3x+6=0`
`<=>x(x-2)-3(x-2)=0`
`<=>(x-2)(x-3)=0`
`<=>` \(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
Vậy `S={2;3}`
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8230
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e/ \(x^2-2x+1=4\\↔(x-1)^2-4=0\\↔(x-1-2)(x-1+2)=0\\↔(x-3)(x+1)=0\\↔\left[\begin{array}{1}x-3=0\\x+1=0\end{array}\right.\\↔\left[\begin{array}{1}x=3\\x=-1\end{array}\right.\)
Vậy \(S=\{3;-1\}\)
g/ \(2x^2-3x+1=0\\↔2x^2-2x-x+1=0\\↔2x(x-1)-(x-1)=0\\↔(2x-1)(x-1)=0\\↔\left[\begin{array}{1}2x-1=0\\x-1=0\end{array}\right.\\↔\left[\begin{array}{1}x=\dfrac{1}{2}\\x=1\end{array}\right.\)
Vậy \(S=\{\dfrac{1}{2};1\}\)
h/ \(x^2-5x+6=0\\↔x^2-2x-3x+6=0\\↔x(x-2)-3(x-2)=0\\↔(x-3)(x-2)=0\\↔\left[\begin{array}{1}x-3=0\\x-2=0\end{array}\right.\\↔\left[\begin{array}{1}x=3\\x=2\end{array}\right.\)
Vậy \(S=\{3;2\}\)
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