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C
26 tháng 9 2019
a) \(3x^2-9x+30=3\left(x^2-3x+10\right)\)
b) \(3x^2-5x-2=3x^2-6x+x-2\)
\(=3x\left(x-2\right)+\left(x-2\right)=\left(3x+1\right)\left(x-2\right)\)
c) \(x^4+4y^4\)
\(=x^4+4y^4+2x^2y^2+2x^2y^2-4x^2y^2+4xy^3-4xy^3+2x^3y-2x^3y\)
\(=\left(4y^4-4xy^3+2x^2y^2\right)+\left(4xy^3-4x^2y^2+2x^3y\right)\)
\(+\left(2x^2y^2-2x^3y+x^4\right)\)
\(=2y^2\left(2y^2-2xy+x^2\right)+2xy\left(2y^2-2xy+x^2\right)\)
\(+x^2\left(2y^2-2xy+x^2\right)\)
\(=\left(2y^2+2xy+x^2\right)\left(2y^2-2xy+x^2\right)\)
d) \(x^5+x+1\)
\(=x^5+x+1+x^4-x^4+x^3-x^3+x^2-x^2\)
\(=\left(x^5-x^4+x^2\right)+\left(x^4-x^3+x\right)+\left(x^3-x^2+1\right)\)
\(=x^2\left(x^3-x^2+1\right)+x\left(x^3-x^2+1\right)+\left(x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
C
30 tháng 10 2019
1) \(x^2-2x-4y^2-4y\)
\(=x^2-2x-4y^2-4y+2xy-2xy\)
\(=\left(-4y^2+2xy-4y\right)-\left(2xy-x^2+2x\right)\)
\(=2y\left(-2y+x-2\right)+x\left(-2y+x-2\right)\)
\(=\left(2y+x\right)\left(-2y+x-2\right)\)
KN
21 tháng 10 2019
b) \(x^3-3x^2+2\)
\(=x^3-2x^2-x^2+2\)
\(=x^2\left(x-2\right)-\left(x-2\right)\left(x+2\right)\)
\(=\left(x^2-x-2\right)\left(x-2\right)\)
c) \(x^4y^4+64\)
\(=x^4y^4+16x^2+64-16x^2\)
\(=\left(x^2y^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2y^2-4x+8\right)\left(x^2y^2+4x+8\right)\)
d) \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+1\)
\(=x^6\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x^2+x+1\right)\left[x^6-\left(x-1\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^6-x^4-x+x^3-1\right]\)
4 tháng 8 2017
a )\(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-2y-2\right)\left(x+2y\right)\)
b )\(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+2x+2\right)\left(x^2-2\right)\)
c ) \(x^2\left(1-x^2\right)-4-4x^2=x^2-x^4-4-4x^2\)
\(=x^2-\left(x^2+2\right)^2=\left(x-x^2-2\right)\left(x^2+x+2\right)\)
29 tháng 8 2018
mk viết đáp án, ko biết biến đổi ib mk
a) \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)
b) \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)
c) \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d) \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)
21 tháng 7 2018
\(x^2-4y^2+4y-1=x^2-\left(2y-1\right)^2=\left(x+2y-1\right)\left(x-2y+1\right)\)
21 tháng 7 2018
\(x^4+3x^3-9x-9\)
\(=x^4-9+3x^3-9x\)
\(=\left(x^2-3\right)\left(x^2+3\right)+3x\left(x^2-3\right)\)
\(=\left(x^2-3\right)\left(x^2+3+3x\right)\)
5 tháng 8 2016
a) \(45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)-\left(9x-45\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)
5 tháng 8 2016
\(a,45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)+\left(45-9x\right)\)
\(=x^2\left(x-5\right)+9\left(5-x\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)\)
\(=\left(x-5\right)\left(x^2-3^2\right)\)
\(=\left(x-5\right)\left(x+3\right)\left(x-3\right)\)
\(c,2x^2+3x-5\)
\(=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(2x+5\right)\)
\(e,\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\left(2\right)\)
(1)\(\Leftrightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\left(3\right)\)
Thay (3) vào (2),ta được:\(\left(x^2+10x+20\right)^2\)
CD
5
10 tháng 7 2017
a) \(x^3+6x^2=x^2\left(x+6\right)\)
b) \(x^4+4y^4=1\left(x^4+4y^4\right)\)
\(x^4+3x+4\)
< không thể phân tích >
mk nghi đề là :
\(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x^2-4\left(x-1\right)\right)\right]\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\left(x-2\right)^2\)
Hoặc : \(x^4+3x^2-4\)
\(=x^4+4x^2-x^2-4\)
\(=x^2\left(x^2+4\right)-\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x^2-1\right)\)
\(=\left(x^2-4\right)\left(x-1\right)\left(x+1\right)\)
_bạn xem lại đề nha_
_Tử yên_
\(x^4+4y^4\)
\(=\left(x^2\right)^2+\left(2y^2\right)^2\)
\(=\left(x^2\right)^2+\left(2.2x^2y^2\right)+\left(2y^2\right)^2-\left(2.2x^2y^2\right)\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)
_Tử yên_