phân tích đa thức thành nhân tử:
a) 3(a-b)+2(a-b)2
b) (a+2)2-(4-a2)
c) (9a2-4b2)+4b-1
K
Khách
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NT
3
3 tháng 7 2017
\(\left(a+b+c\right)^2+\left(a-b+c\right)^2-4b^2\)
\(=2a^2+2b^2+2c^2+2ab+2ac+2bc-2ab-2bc+2ac-4b^2\)
\(=2a^2-2b^2+2c^2+4ac\)
\(=2\left[\left(a^2+2ac+c^2\right)-b^2\right]=2\left[\left(a+c\right)^2-b^2\right]\)
\(=2\left(a+c-b\right)\left(a+b+c\right)\)
NM
1
17 tháng 6 2016
\(\left(a+b+c\right)^2+\left(a-b+c\right)^2-4b^2=\left[\left(a+b+c\right)^2-4b^2\right]+\left(a-b+c\right)^2=\left(a-b+c\right)\left(a+3b+c\right)+\left(a-b+c\right)^2=\left(a-b+c\right)\left(2a+2b+2c\right)=2\left(a+b+c\right)\left(a-b+c\right)\)
T
4 tháng 9 2023
\(a,a^2-2a-4b^2-4b\)
\(=\left(a^2-4b^2\right)-\left(2a+4b\right)\)
\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)
\(=\left(a+2b\right)\left(a-2b-2\right)\)
\(b,x^3-2x^2+4x-8\)
\(=x^2\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
\(c,x^3+36x-12x^2\)
\(=x^3-6x^2-6x^2+36x\)
\(=x^2\left(x-6\right)-6x\left(x-6\right)\)
\(=\left(x-6\right)\left(x^2-6x\right)\)
\(=x\left(x-6\right)^2\)
\(d,5a^2+3\left(a+b\right)^2-5b^2\)
\(=\left(5a^2-5b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a^2-b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a-b\right)\left(a+b\right)+3\left(a+b\right)^2\)
\(=\left(a+b\right)\left[5\left(a-b\right)+3\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(5a-5b+3a+3b\right)\)
\(=\left(a+b\right)\left(8a-2b\right)\)
\(=2\left(a+b\right)\left(4a-b\right)\)
\(e,x^3-3x^2+3x-1-y^3\)
\(=\left(x^3-3x^2+3x-1\right)-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-y-1\right)\left(x^2-2x+1+xy-y+y^2\right)\)
\(=\left(x-y-1\right)\left(x^2+y^2-xy-y+1\right)\)
#Urushi☕
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
4 tháng 9 2023
\(c.\\ x^3+36x-12x^2\\ =x\left(x^2-12x+36\right)\\ =x.\left(x^2-2.x.6+6^2\right)\\ =x.\left(x-6\right)^2\\ ---\\ d.\\ 5a^2+3\left(a+b\right)^2-5b^2\\ =\left(5a^2-5b^2\right)+3\left(a+b\right)^2\\ =5.\left(a^2-b^2\right)+3.\left(a+b\right)\left(a+b\right)\\ =5\left(a+b\right)\left(a-b\right)+3\left(a+b\right)\left(a+b\right)\\ =\left(a+b\right)\left(5a-5b+3a+3b\right)\\ =\left(a+b\right)\left(8a-2b\right)\\ =2\left(a+b\right)\left(4a-b\right)\)
\(e.\\ x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right).y+y^2\right]\\ =\left(x-y-1\right).\left[\left(x^2-2x+1\right)+y\left(x+y-1\right)\right]\)
KN
9 tháng 9 2019
a) \(\left(6x-1\right)^2-\left(3x+2\right)^2\)
\(=\left(6x-1+3x+2\right)\left(6x-1-3x-2\right)\)
\(=\left(9x+1\right)\left(3x-3\right)\)
\(=3\left(9x+1\right)\left(x-1\right)\)
b) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)
\(=\left(6x+9+2x+2\right)\left(6x+9-2x-2\right)\)
\(=\left(8x+11\right)\left(4x+7\right)\)
c) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)
\(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)
\(=-\left[\left(b+c\right)^2-a^2\right]\left(b^2-2bc+c^2-a^2\right)\)
\(=-\left(b+c-a\right)\left(b+c+a\right)\left[\left(b-c\right)^2-a^2\right]\)
\(=-\left(b+c-a\right)\left(b+c+a\right)\left(b-c-a\right)\left(b-c+a\right)\)
KN
9 tháng 9 2019
d) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)
\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)
\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-3^2\right]\)
\(=\left(a+b+1\right)\left(a+b-1\right)\left(a-b-3\right)\left(a-b+3\right)\)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
3 tháng 12 2016
a) \(\left(2x+5\right)^2\)\(-\left(x-9\right)^2\)
=\(\left(2x+5+x-9\right).\left(2x+5-x+9\right)\)
=\(\left(3x-4\right).\left(x+14\right)\)
a: \(3\cdot\left(a-b\right)+2\left(a-b\right)^2\)
\(=\left(a-b\right)\left(3+2a-2b\right)\)
b:
\(\left(a+2\right)^2-\left(4-a^2\right)\)
\(=\left(a+2\right)^2+\left(a-2\right)\left(a+2\right)\)
\(=\left(a+2\right)\left(a+2+a-2\right)\)
\(=2a\left(a+2\right)\)
c: \(9a^2-4b^2+4b-1\)
\(=\left(3a\right)^2-\left(4b^2-4b+1\right)\)
\(=\left(3a\right)^2-\left(2b-1\right)^2\)
\(=\left(3a-2b+1\right)\left(3a+2b-1\right)\)
\(a,3\left(a-b\right)+2\left(a-b\right)^2=\left(a-b\right)\left[3+2\cdot\left(a-b\right)\right]=\left(a-b\right)\left(3+2a-2b\right)\)
\(b,\left(a+2\right)^2-\left(4-a^2\right)\\ =\left(a+2\right)^2-\left(2^2-a^2\right)\\ =\left(a+2\right)^2-\left(2-a\right)\left(a+2\right)\\ =\left(a+2\right)\left(a+2-2+a\right)\\ =\left(a+2\right)\cdot2a\)
\(c,\left(9a^2-4b^2\right)+4b-1\\ =9a^2-\left(4b^2-4b+1\right)\\ =\left(3a\right)^2-\left(2b-1\right)^2\\ =\left(3a-2b+1\right)\left(3a+2b-1\right)\)