`1)` `x^3+5x^2-4x-20=0`
`=>x^2(x+5)-4(x+5)=0`
`=>(x+5)(x^2-4)=0`
`=> (x+5)(x+2)(x-2)=0`
`=>\(\left[ \begin{array}{l}x=-5\\x=-2\\x=2\end{array} \right.\)
`2)` `x^3+3x^2-3x-1=?`
Thiếu đề nha bn
VD: `x^3+3x^2-3x-1=2`
`=>x^3+3x^2-3x-3=0`
`=>x^2(x+3)-3(x+3)=0`
`=>(x+3)(x^2-3)=0`
`=>`\(\left[ \begin{array}{l}x+3=0\\x^2-3=0\end{array} \right.\)
`=>`\(\left[ \begin{array}{l}x=-3\\x=±\sqrt{3}\end{array} \right.\)
Đáp án + Giải thích các bước giải:
1)
`x^3+5x^2-4x-20=0`
`=>(x+5)(x-2)(x+2)=0`
`=>x+5=0` hoặc `x-2=0` hoặc `x+2=0`
`=>x=-5` hoặc `x=2` hoặc `x=-2`
Vậy `S={-5;2;-2}`
2)
Thiếu đề ạ (không có dấu `=`)
Cho A= (4n+3)^2 - 25. Chứng minh A chia hết cho 8 với mỗi n là số nguyên.
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