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3 câu trả lời 294
1.
a. x3-3x2+3x-1-y3=(x-1)3-y3 =(x-1-y)[(x-1)2+(x-1)y+y2]b. x3-2x2+4x-8= (x-2)3
Bài $1$:
`a)`
`x^3-3x^2+3x-1-y^3`
`=(x^3-3x^2+3x-1)-y^3`
`=(x-1)^3-y^3`
`=(x-1-y)[(x-1)^2+(x-1).y+y^2]`
`=(x-1-y)(x^2-2x+1+xy-y+y^2)`
`b)`
`x^3-2x^2+4x-8`
`=(x^3-2x^2)+(4x-8)`
`=x^2(x-2)+4(x-2)`
`=(x-2)(x^2+4)`
Bài $2$ :
`a)` `8x^3-50x=0`
`⇔` `2x(4x^2-25)=0`
`⇔` `2x(2x-5)(2x+5)=0`
`⇔` \(\left[ \begin{array}{l}2x=0\\2x-5=0\\2x+5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\2x=5\\2x=-5\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{array} \right.\)
Vậy `S={0;5/2;-5/2}`
`b)` `6x^2-15x-(2x+5)(2x-5)=0`
`⇔` `6x^2-15x-4x^2+25=0`
`⇔` `2x^2-15x+25=0`
`⇔` `2x^2-10x-5x+25=0`
`⇔` `2x(x-5)-5(x-5)=0`
`⇔` `(2x-5)(x-5)=0`
`⇔` \(\left[ \begin{array}{l}2x-5=0\\x-5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}2x=5\\x=5\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=5\end{array} \right.\)
Vậy `S={5/2;5}`
`c)` `3x^2-5x-2=0`
`⇔` `3x^2-6x+x-2=0`
`⇔` `3x(x-2)+(x-2)=0`
`⇔` `(x-2)(3x+1)=0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\3x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\3x=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=-\dfrac{1}{3}\end{array} \right.\)
Vậy `S={2;-1/3}`
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