\(\left(x+1\right)^2=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow x^2+2x+1-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2+2x+1-4x^2+8x-4=0\)
\(\Leftrightarrow-3x^2+10x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{3};3\right\}\)
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`(x+1)^2=4(x^2-2x+1)`
`<=>(x+1)^2-[2(x-1)]^2=0`
`<=>(x+1-2x+2)(x+1+2x-2)=0`
`<=>(-x+3)(3x-1)=0`
`<=>-x+3=0` hoặc `3x-1=0`
`<=>x=3` hoặc `x=1/3`
Vậy `S={1/3;3}`
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