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\(a^3b-ab^3+a^2+2ab+b^2\)
\(=\left(a^3b-ab^3\right)+\left(a^2+2ab+b^2\right)\)
\(=ab\left(a^2-b^2\right)+\left(a+b\right)^2\)
\(=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\)
\(=\left(a+b\right)\left[ab\left(a-b\right)+\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(a^2b-ab^2+a+b\right)\)
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
Bài 1: 4a2-4ab+b2-9a2b2
=(2a)2-2.2a.b+b2-(3ab)2
=(2a-b)2-(3ab)2
=(2a-b-3ab)(2a-b+3ab)
a/ (4a2-4ab+b2)-9a2b2
= (2a-b)2-(3ab)2
= (2a-b-3ab) (2a-b+3ab)
1/ phân tích thành nhân tử ;
= C2-( a +b )2=( c-a -b ) . ( c+a +b )
\(3x^2-3y^2\)
\(=3\left(x^2-y^2\right)\)
\(=3\left(x-y\right)\left(x+y\right).\)
Merry Chrismas.
\(3x^2-3y^2=3\left(x^2-y^2\right)=3\left(x+y\right)\left(x-y\right)\)
a)\(2a^2-3ab+b^2\)
=\(a^2+a^2-2ab-ab+b^2\)
=\(\left(a-b\right)^2+a\left(a-b\right)\)
=\(\left(a-b\right)\left(2a-b\right)\)
b)\(x^2-7x-30\)
=\(x^2-10x+3x-30\)
=\(x\left(x-10\right)+3\left(x-10\right)\)
=\(\left(x-10\right)\left(x+3\right)\)
c)\(6a^2-5ab-6b^2\)
=\(6a^2-9ab+4ab-6b^2\)
=\(3a\left(2a-3b\right)+2b\left(2a-3b\right)\)
=\(\left(2a-3b\right)\left(3a+2b\right)\)
d)\(a^4+a^2+1\)
=\(a^4+2a^2-a^2+1\)
=\(\left(a^2+1\right)^2-a^2\)
=\(\left(a^2+1-a\right)\left(a^2+1+a\right)\)
e)\(x^3+6x^2+11x+6\)
=\(x\left(x^2+6x+9+2\right)+6\)
\(=x\left(\left(x+3\right)^2+2\right)+6\)
=\(x\left(x+3\right)^2+2x+6\)
=\(x\left(x+3\right)^2+2\left(x+3\right)\)
=\(\left(x+3\right)\left(x^2+3x+2\right)\)
a, Sửa đề :
\(a^2+b^2-ac+2ab-bc\)
\(=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)
b, \(\frac{1}{4}a^2b-bc^4=b\left(\frac{1}{4}a^2-c^4\right)=b\left(\frac{1}{2}a-c^2\right)\left(\frac{1}{2}a+c^2\right)\)
bạn tham khảo 1 số bài rồi tự làm nhé
a) 3x−3+5(1−x)
=3x−3+5−5x
=3x−5x+2
=x(3−5)+2
=−2x+2
=2(1−x)
b) 12a2−3ab+8ac−2bc
=3a(4a−b)+2c(4a−b)
=(4a−b)(3a+2c)
c) x2−25+y2−2xy
=x2−2xy+y2−25
=(x−y)2−52
=(x−y−5)(x−y+5)
Bài 1:
a: \(4a^2-6b=2\left(2a^2-3b\right)\)
b: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
Bài 1:
a) \(4a^2-6b=2\left(a^2-3b\right)\)
b) \(=mn\left(m^2-2mn-1\right)\)
Bài 2:
a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)
b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)