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a) Ta có: \(a^2-b^2-5a+5b\)
\(=\left(a-b\right)\left(a+b\right)-5\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b-5\right)\)
b) Ta có: \(a^2-b^2-3ab^2-3a^2b\)
\(=\left(a-b\right)\left(a+b\right)-3ab\left(a+b\right)\)
\(=\left(a+b\right)\left(a-b-3ab\right)\)
1. = 2(a+b)
2. =2(a-b)
3.=2(a+2b-3c)
4.=3(a-2b-3c)
5.=-4(a+2b+3c)
6.=-5(x+2xy+3y)
7.=-7(a+2ab+3b)
8.=6(xy-2x-3y)
9.=8(xy-3y+2x)
10.=9(ab-2a+1)
Lần sau bn cần ghi đề rõ ràng hơn
11.=x(y-1)
12.=a(x+1)
13.=m(x+y+1)
14.=-a(x+y+1)
15.=-a(x2+x+1)
16.=-2a(x+2y)
17.=2a(x-y+1)
18.=2(2ax-ay-2)
19.=5a(1-2x-3)
20.=-2ab(a+2b+3)
a) \(6a^2b^2c-4ab^2c^2+12a^2bc^2\)
\(=2abc\left(3ab-2bc+6ac\right)\)
b)\(x^2\left(x-y\right)-y\left(y-x\right)\)
\(=x^2\left(x-y\right)+y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+y\right)\)
Phân tích đa thức thành nhân tử
a. 3ab ( x+ y) - 6ab ( y+ x)
=( x + y) ( 3ab - 6ab )
= ( x +y ) ( - 3ab)
b.7a (x - 3)+a2(x2 - 9)
=7a( x- 3) + a2 ( x2 - 32)
=7a ( x - 3 ) + a2 ( x- 3 ) ( x+3 )
= ( x- 3) . 7a + a2 ( x + 3)
= ( x- 3) ( 7a +a2x + 3a2)
c. 34 (x + y) -x -y
= 34 ( x+ y) - ( x+y)
=(x +y ) ( 34 - 1) = 33 ( x+ y)
d. 25 x4 - 942
=( 5x2 )2 - 942
=( 5x2 - 94 ) ( 5x2+94)
e.( 5a - b )2 - ( 2a +3b)2
=( 5a -b -2a - 3b) (5a -b + 2a + 3b)
=(3a - 4b) (7a+ 2b)
k. 22 -3a - b2 +3b
=( 22 - b2 ) + ( -3a +3b)
=( 2-b) (2+b) + 3( -a +b)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(a,15a^2b^3+5a^3b^2=5a^2b^2\left(3b+a\right)\\ b,x^2-2x+1-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
a) 15a2b3+5a3b2=5a2b2(3b+a)
b) x2-2x+x-y2=( x2-y2)-(2x+x)
=(x-y)(x+y)-x(2-1)
=(x-y)(x+y)-x3
a) \(5a-10ax-15a\)
\(=5a+5ax-15ax-15a\)
\(=5a\left(1+x\right)-15a\left(x+1\right)\)
\(=\left(x+1\right)\left(5a-15a\right)\)
\(=-10a\left(x+1\right)\)
b) \(-2a^2b-4ab^2-6ab\)
\(=-\left(2a^2b+4ab^2+6ab\right)\)
\(=-2ab\left(a+2b+3\right)\)