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a) \(x^2-2x-15\)
\(\Leftrightarrow x^2-2x+1-16\)
\(\Leftrightarrow\left(x-1\right)^2-4^2\)
\(\Leftrightarrow\left(x-5\right)\left(x-3\right)\)
\(a,x^2-2x-15=\left(x^2-2x+1\right)-16.\)
\(=\left(x-1\right)^2-4^2\)
\(=\left(x-5\right)\left(x+3\right)\)
1/ x3 + x2y - x2z -xyz
= x2(x + y) - xz(x + y)
= (x + y) (x2 - xz)
= x (x + y) (x - z)
2/ 3x3y - 18x2y2 + 27xy3
= 3xy(x2 - 6xy + y2)
\(2,25x^2-12x-13\)
\(=25x^2-25x+13x-13\)
\(=25x\left(x-1\right)+13\left(x-1\right)\)
\(=\left(x-1\right)\left(25x+13\right)\)
\(3,2y^2-3y-5\)
\(=2y^2+2y-5y-5\)
\(=2y\left(y+1\right)-5\left(y+1\right)\)
\(=\left(y+1\right)\left(2y-5\right)\)
Còn bài 1 mik đang nghĩ, khi nào biết mik trả lời nha!!!
Chúc bn học giỏi!!!
a) \(A=x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2-1+3x-3y-3\)
\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1+3\right)\)
\(=\left(x-y-1\right)\left(x-y+4\right)\)
Phân tích đa thức thành nhân tử:
\(3x^2-12x^2y^2+3y^2+6xy\)
\(=3\left(x^2-4x^2y^2+y^2+2xy\right)\)
\(=3\left[\left(x^2+2xy+y^2\right)-\left(2xy\right)^2\right]\)
\(=3\left[\left(x+y\right)^2-\left(2xy\right)^2\right]\)
\(=3\left(x+y-2xy\right)\left(x+y+2xy\right)\)
\(3y^3+6xy^2+3x^2y=3y\left(y^2+2xy+x^2\right)=3y\left(x+y\right)^2\)
\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+3x\right)\)
\(=\left(x-1\right)\left(x^2+4x+1\right)\)
Tham khảo nhé~
1) \(\left(3x^2-3y^2\right)-\left(12x-12y\right)\)
\(=3xy\left(x-y\right)-12\left(x-y\right)\)
\(=\left(3xy-12\right)\left(x-y\right)\)
2) \(4x^3+4xy^2+8x^2y-16x\)
\(=\left(4x^3-16x\right)+\left(4xy^2+8x^2y\right)\)
\(=4x\left(x^2-4\right)+4xy\left(y+2x\right)\)
Ta có : 3x2 - 3y2 - 12x + 12y
= (3x2 - 3y2) - (12x - 12y)
= 3(x2 - y2) - 12(x - y)
= 3(x - y)(x + y) - 4.3.(x - y)
= 3(x - y)(x + y - 4)
a) \(12x^5y+24x^4y^2+12x^3y^3\)
\(=12x^3y\left(x^2+2xy+y^2\right)\)
\(=12x^3y\left(x+y\right)^2\)
b) \(x^2-2xy-4+y^2\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
g) \(12xy-12xz+3x^2y-3x^2z\)
\(=12x\left(y-z\right)+3x^2\left(y-z\right)\)
\(=3x\left(4+x\right)\left(y-z\right)\)
e) \(16x^2-9\left(x^2+2xy+y^2\right)\)
\(=\left(4x\right)^2-\left[3\left(x+y\right)\right]^2\)
\(=\left(4x-3\left(x+y\right)\right)\left(4x+3\left(x+y\right)\right)\)
\(=\left(x+y\right)\left(7x+y\right)\)
d) làm tương tự như phần g chỉ khác là phải nhóm( nhóm xen kẽ), phần f cũng vậy
a) \(x^2y+2xy+y=y\left(x^2+2x+1\right)=y\left(x+1\right)^2\)
b) \(4x^2-4xy-6y^2+6xy=4x\left(x-y\right)+6y\left(x-y\right)=\left(x-y\right)\left(4x+6y\right)\)
\(=2\left(x-y\right)\left(2x+3y\right)\)
c) \(18x^5y+18x^3y-2x^3y^5-2xy^5=18x^3y\left(x^2+1\right)-2xy^5\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(18x^3y-2xy^5\right)=2xy\left(x^2+1\right)\left(9x^2-y^4\right)=2xy\left(x^2+1\right)\left(3x-y^2\right)\left(3x+y^2\right)\)
d)
d) \(-12x^5-12x^3y-3xy^2+36x^4+36x^2y+9y^2=-3x\left(4x^4+4x^2y+y^2\right)+9y\left(4x^4+4x^2y+y^2\right)\)\(=\left(4x^4+4x^2y+y^2\right)\left(9-3x\right)\)
a; 18\(x^2\)y3 - 12\(x^3\)y2
= 3\(x^2\)y2.(6y - 4\(x\))
b; \(x^2\) - y2 + 3\(x\) + 3y
= (\(x^2\) - y2) + 3(\(x+y\))
= (\(x-y\))(\(x+y\)) + 3(\(x+y\))
= (\(x+y\))(\(x-y\) + 3)