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12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
TM
20 tháng 7 2016
a)\(36-4a^2+20ab-25b^2=6^2-\left(4a^2-20ab+25b^2\right)\)
\(=6^2-\left[\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right]\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
b)\(a^3+3a^2+3a+1-27b^3=\left(a+1\right)^3-\left(3b\right)^3\)(chỗ này mình sửa 27b2 thành 27b3 vì mình nghĩ nhầm đề)
\(=\left(a+1-3b\right)\left[\left(a+1\right)^2+\left(a+1\right)3b+\left(3b\right)^2\right]\)
\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)
c)\(x^3+3x^2+3x+1-3x^2-3x=\left(x+1\right)^3-3x\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\)
LC
28 tháng 9 2018
c, \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2[x^2\left(x-1\right)\left(x+1\right)+2\left(x+1\right)]\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
\(=x^2\left(x+1\right)[x^2\left(x+1\right)-2x\left(x+1\right)+2\left(x+1\right)]\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
LC
28 tháng 9 2018
d,
\(2x^3-x^2-1\)
\(=2x^3-2x^2+x^2-x+x-1\)
\(=2x^2\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(2x^2+x+1\right)\)
NB
0
NT
2
26 tháng 6 2021
`a)3x-3a+yx-ya`
`=3(x-a)+y(x-a)`
`=(x-a)(y+3)`
`b)x^2-9-4(x+3)`
`=(x-3)(x+3)-4(x+3)`
`=(x+3)(x-3-4)`
`=(x+3)(x-7)`
PK
1
25 tháng 10 2020
a) \(\frac{1}{3}x^2y+\frac{1}{6}xy^2-\frac{1}{9}xy\)
\(=xy\left(\frac{1}{3}x+\frac{1}{6}y-\frac{1}{9}\right)\)
b) \(a^3+3a^2+3a-7\)
\(=\left(a^3+3a^2+3a+1\right)-8\)
\(=\left(a+1\right)^3-2^3\)
\(=\left(a+1-2\right)\left[\left(a+1\right)^2+2\left(a+1\right)+2^2\right]\)
\(=\left(a-1\right)\left(a^2+2a+1+2a+2+4\right)\)
\(=\left(a-1\right)\left(a^2+4a+7\right)\)
c) \(2x\left(2x-1\right)-2x^2\)
\(=4x^2-2x-2x^2\)
\(=2x^2-2x=2x\left(x-1\right)\)
NH
5
MT
8 tháng 2 2016
Ta có:A= 3a+3b-a^2-ab
=>A= (3a-a^2)+(3b-ab)
=>A= a(3-a)+b(3-a)
=>A= (a+b)(3-a)
PH
0
sửa đề : \(a^3-3a^2+3a-1-27b^3\)
giải:
\(a^3+3.a^2.\left(-1\right)+3a\left(-1\right)^2+\left(-1\right)^3-\left(3b\right)^3\\ =\left(a-1\right)^3-\left(3b\right)^3\\ =\left(a-1-3b\right)\left(a^2-2a+1+3ab+3b+9b^2\right)\)