Bài 4:
a, `2(x+5)-x^2-5x=0`
`<=>2x+10-x^2-5x=0`
`<=>2(x+5)-x(x+5)=0`
`<=>(x+5)(2-x)=0`
`<=>`\(\left[ \begin{array}{l}x+5=0\\2-x=0\end{array} \right.\)`<=>`\(\left[ \begin{array}{l}x=2\\x=-5\end{array} \right.\)
Vậy `S={-5;2}`
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b, `x^2+3x-5=0`
`<=>x^2+3x+9/4-29/4=0`
`<=>(x+3/2)^2=29/4`
`<=>(x+3/2)^2=(±(sqrt29)/2)`
`<=>`\(\left[ \begin{array}{l}x+\dfrac32=\dfrac{-\sqrt{29}}{2}\\x+\dfrac32=\dfrac{29}{2}\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=\dfrac{-3+\sqrt{29}}{2}\\x=\dfrac{-3+\sqrt{29}}{2}\end{array} \right.\)
Vậy `S={(-3±sqrt29)/2}`
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Bài 5:
a, ĐKXĐ: `x\ne0;x\ne±2`
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b, `A=x^2/(x^3-4x)+3/(6-3x)+1/(x+2)`
`=x^2/(x(x-2)(x+2))+3/(3(2-x))+1/(x+2)`
`=x/((x-2)(x+2))+1/(2-x)+1/(x+2)`
`=x/((x-2)(x+2))-1/(x-2)+1/(x+2)`
`=x/((x-2)(x+2))-(x+2)/((x-2)(x+2))+(x-2)/((x-2)(x+2))`
`=(x-x-2+x-2)/((x-2)(x+2))`
`=(x-4)/((x-2)(x+2))`
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c, Với `x=3`, ta có:
`A=(3-4)/((3-2)(3+2))=(-1)/5`
