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Khách
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12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
NK
3
29 tháng 8 2021
phân tích đa thức thành nhân tử
a^2(b-c)+b^2(c-a)+c^2(a-b)
= -(b-a)(c-a)(c-b)
nha bạn
TD
30 tháng 8 2021
a2(b-c)+b2(c-a)+c2(a-b)
=a2b-a2c+b2c-b2a+c2(a-b)
=(a2b-b2a)-(a2c-b2c)+c2(a-b)
=ab(a-b)+c(a2-b2)+c2(a-b)
=ab(a-b)+c(a-b)(a+b)+c2(a-b)
=(a-b)(ab+ac+bc+c2)
=(a-b)[(ab+bc)+(ac+c2)]
=(a-b)[b(a+c)+c(a+c)]
=(a-b)(a+c)(b+c)
11 tháng 11 2021
\(\left(a+b-a+b\right)\left(a+b+a-b\right)=2b\cdot2a=4ab\)
GL
0
NT
3
3 tháng 7 2017
\(\left(a+b+c\right)^2+\left(a-b+c\right)^2-4b^2\)
\(=2a^2+2b^2+2c^2+2ab+2ac+2bc-2ab-2bc+2ac-4b^2\)
\(=2a^2-2b^2+2c^2+4ac\)
\(=2\left[\left(a^2+2ac+c^2\right)-b^2\right]=2\left[\left(a+c\right)^2-b^2\right]\)
\(=2\left(a+c-b\right)\left(a+b+c\right)\)
TT
1
6 tháng 8 2022
\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)
\(=\left(a^2b-b^2a\right)-\left(a^2c-b^2c\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ca-cb+c^2\right)\)
\(=\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
TG
1
18 tháng 10 2018
\(\left(a^2-b^2\right)+\left(a^3+b^3\right)-a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a-b\right)+\left(a+b\right)\left(a^2-ab+b^2\right)-a^2b^2\left(a+b\right)\)
\(=\left(a+b\right)\left(a-b+a^2+b^2-ab-a^2b^2\right)\)
\(=\left(a+b\right)\left[b^2\left(1-a^2\right)+a\left(1+a\right)-b.\left(1+a\right)\right]\)
\(=\left(a+b\right)\left(a+1\right)\left(b^2+a-b\right)\)
NN
2
26 tháng 6 2019
\(a\left(b^2-c^2\right)-b\left(a^2-c^2\right)+c\left(a^2-b^2\right)\)
\(=ab^2-ac^2-ba^2+bc^2+ca^2-cb^2\)
\(=\left(ab^2-ac^2-bc^2\right)-\left(ba^2-bc^2-ca^2\right)\)
\(=a\left(b^2-c^2\right)-bc^2-a^2\left(b-c\right)+bc^2\)
\(=a\left(b^2-c^2\right)-a^2\left(b-c\right)\)
\(=a\left(b-c\right)\left(b+c\right)-a^2\left(b-c\right)\)
\(=\left(b+c\right)\left[a\left(b-c\right)-a^2\right]\)
\(=\left(b+c\right)\left(ab-ac-a^2\right)\)
26 tháng 6 2019
\(a\left(b^2-c^2\right)-b\left(a^2-c^2\right)+c\left(a^2-b^2\right)\)
\(=c\left(a^2-b^2\right)+a\left(b^2-c^2\right)+b\left(c^2-a^2\right)\)
\(=-c\left[\left(b^2-c^2\right)+\left(c^2-a^2\right)\right]+a\left(b^2-c^2\right)+b\left(c^2-a^2\right)\)
\(=\left(a-c\right)\left(b^2-c^2\right)+\left(b-c\right)\left(c^2-a^2\right)\)
\(=\left(a-c\right)\left(b-c\right)\left(b+c\right)+\left(b-c\right)\left(c-a\right)\left(c+a\right)\)
\(=\left(a-c\right)\left(b-c\right)\left(b-a\right)\)
\(=\left(a+b-a+b\right)\left(a+b+a-b\right)=2b.2a=4ab\)