phân tích đa thức thành nhân tử:phương pháp 2 (a+b)^3-c^3 125-(x+2)^3 (x+3)^3-8 x^6-1 X^6+1 (x-y)^3-(x+y)^3 15/08/2021 Bởi Maya phân tích đa thức thành nhân tử:phương pháp 2 (a+b)^3-c^3 125-(x+2)^3 (x+3)^3-8 x^6-1 X^6+1 (x-y)^3-(x+y)^3
Đáp án: \(a,\ (a+b)^3-c^3=(a+b-c)(a^2+2ab+b^2+ac+bc+c^2)\\ b,\ 125-(x+2)^3=(7-x)(39+9x+x^2)\\ c,\ (x+3)^3-8=(x+1)(x^2+8x+19)\\ d,\ x^6-1=(x-1)(x+1)(x^4+x^2+1)\\ e,\ x^6+1=(x^2+1)(x^4-x^2+1)\\ f,\ (x-y)^3-(x+y)^3=(-2y)(3x^2+y^2)\) Giải thích các bước giải: \(a,\ (a+b)^3-c^3\\ =(a+b-c)\left [(a+b)^2+(a+b).c+c^2\right ]\\ =(a+b-c)(a^2+2ab+b^2+ac+bc+c^2)\\ b,\ 125-(x+2)^3\\ =5^3-(x+2)^3\\ =(5-x+2)\left [5^2+5.(x+2)+(x+2)^2\right ]\\ =(7-x)(25+5x+10+x^2+4x+4)\\ =(7-x)(39+9x+x^2)\\ c,\ (x+3)^3-8\\ =(x+3)^3-2^3\\ =(x+3-2)\left [(x+3)^2+(x+3).2+2^2\right ]\\ =(x+1)(x^2+6x+9+2x+6+4)\\ =(x+1)(x^2+8x+19)\\ d,\ x^6-1\\ =(x^2)^3-1\\ =(x^2-1)\left [(x^2)^2+x^2.1+1^2\right ]\\ =(x-1)(x+1)(x^4+x^2+1)\\ e,\ x^6+1\\ =(x^2)^3+1^3\\ =(x^2+1)\left [(x^2)^2-x^2.1+1^2\right ]\\ =(x^2+1)(x^4-x^2+1)\\ f,\ (x-y)^3-(x+y)^3\\ =(x-y-x-y)\left [(x-y)^2+(x-y)(x+y)+(x+y)^2\right ]\\ =(-2y)(x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2)\\ =(-2y)(3x^2+y^2)\) chúc em học tốt! Bình luận
`a,(a+b)^3-c^3` `=(a+b-c)[(a+b)^2+(a+b)c+c^2]` `(a+b-c)(a^2+2ab+b^2+ac+bc+c^2)` `b,125-(x+2)^3` `=(5-x-2)[25+5(x+2)+(x+2)^2]` `=(3-x)(25+5x+10+x^2+4x+4)` `=(3-x)(x^2+9x+29)` `c=x^6-1` `=(x^2-1)(x^4+x^2+1)` `=(x-1)(x+1)(x^2+x^2+1)` `d,x^6+1` `=(x^2+1)(x^4-x^2+1)` `c,(x-y)^3-(x+y)^3` `=(x-yx-y)[(x-y)^2+(x-y)(x+y)+(x+y)^2]` `=-2y(x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2)` `=-2y(3x^2+y^2)` Bình luận
Đáp án:
\(a,\ (a+b)^3-c^3=(a+b-c)(a^2+2ab+b^2+ac+bc+c^2)\\ b,\ 125-(x+2)^3=(7-x)(39+9x+x^2)\\ c,\ (x+3)^3-8=(x+1)(x^2+8x+19)\\ d,\ x^6-1=(x-1)(x+1)(x^4+x^2+1)\\ e,\ x^6+1=(x^2+1)(x^4-x^2+1)\\ f,\ (x-y)^3-(x+y)^3=(-2y)(3x^2+y^2)\)
Giải thích các bước giải:
\(a,\ (a+b)^3-c^3\\ =(a+b-c)\left [(a+b)^2+(a+b).c+c^2\right ]\\ =(a+b-c)(a^2+2ab+b^2+ac+bc+c^2)\\ b,\ 125-(x+2)^3\\ =5^3-(x+2)^3\\ =(5-x+2)\left [5^2+5.(x+2)+(x+2)^2\right ]\\ =(7-x)(25+5x+10+x^2+4x+4)\\ =(7-x)(39+9x+x^2)\\ c,\ (x+3)^3-8\\ =(x+3)^3-2^3\\ =(x+3-2)\left [(x+3)^2+(x+3).2+2^2\right ]\\ =(x+1)(x^2+6x+9+2x+6+4)\\ =(x+1)(x^2+8x+19)\\ d,\ x^6-1\\ =(x^2)^3-1\\ =(x^2-1)\left [(x^2)^2+x^2.1+1^2\right ]\\ =(x-1)(x+1)(x^4+x^2+1)\\ e,\ x^6+1\\ =(x^2)^3+1^3\\ =(x^2+1)\left [(x^2)^2-x^2.1+1^2\right ]\\ =(x^2+1)(x^4-x^2+1)\\ f,\ (x-y)^3-(x+y)^3\\ =(x-y-x-y)\left [(x-y)^2+(x-y)(x+y)+(x+y)^2\right ]\\ =(-2y)(x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2)\\ =(-2y)(3x^2+y^2)\)
chúc em học tốt!
`a,(a+b)^3-c^3`
`=(a+b-c)[(a+b)^2+(a+b)c+c^2]`
`(a+b-c)(a^2+2ab+b^2+ac+bc+c^2)`
`b,125-(x+2)^3`
`=(5-x-2)[25+5(x+2)+(x+2)^2]`
`=(3-x)(25+5x+10+x^2+4x+4)`
`=(3-x)(x^2+9x+29)`
`c=x^6-1`
`=(x^2-1)(x^4+x^2+1)`
`=(x-1)(x+1)(x^2+x^2+1)`
`d,x^6+1`
`=(x^2+1)(x^4-x^2+1)`
`c,(x-y)^3-(x+y)^3`
`=(x-yx-y)[(x-y)^2+(x-y)(x+y)+(x+y)^2]`
`=-2y(x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2)`
`=-2y(3x^2+y^2)`