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a) 5x2 - 5xy + 7y - 7x = ( 5x2 - 5xy ) - ( 7x - 7y ) = 5x( x - y ) - 7( x - y ) = ( x - y )( 5x - 7 )
b) x2 - y2 + 2x + 1 = ( x2 + 2x + 1 ) - y2 = ( x + 1 )2 - y2 = ( x - y + 1 )( x + y + 1 )
c) 3x2 + 6xy + 3y2 - 3z2 = 3( x2 + 2xy + y2 - z2 ) = 3[ ( x2 + 2xy + y2 ) - z2 ] = 3[ ( x + y )2 - z2 ] = 3( x + y - z )( x + y + z )
d) ab( x2 + y2 ) + xy( a2 + b2 ) = abx2 + aby2 + a2xy + b2xy
= ( a2xy + abx2 ) + ( aby2 + b2xy )
= ax( ay + bx ) + by( ay + bx )
= ( ay + bx )( ax + by )
\(a^3-a^2x-ay+xy\)
\(=a^2\left(a-x\right)-y\left(a-x\right)\)
\(=\left(a-x\right)\left(a^2-y\right)\)
\(4x^2-y^2+4x+1\)
\(=\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2=\left(2x-y+1\right)\left(2x+y+1\right)\)
\(x^3-x+y^3-y\)
\(=\left(x^3+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
a)a3 - a2x - ay +xy
=(a3 - a2x) - (ay - xy)
=a2(a-x) - y(a-x)
=(a-x).(a2 - y)
a: \(x^4+25x^2+20x-4\)
\(=x^4-5x^3+2x^2+5x^3-25x^2+10x-2x^2+10x-4\)
\(=x^2\left(x^2-5x+2\right)+5x\left(x^2-5x+2\right)-2\left(x^2-5x+2\right)\)
\(=\left(x^2-5x+2\right)\left(x^2+5x-2\right)\)
b: \(=x^4-6x^2-x^2+9\)
\(=\left(x^2-3\right)^2-x^2\)
\(=\left(x^2-x-3\right)\left(x^2+x-3\right)\)
c: \(=abx^2+aby^2-a^2xy-b^2xy\)
\(=\left(abx^2-b^2xy\right)+\left(aby^2-a^2xy\right)\)
\(=xb\left(ax-by\right)+ay\left(by-ax\right)\)
\(=\left(ax-by\right)\cdot\left(xb-ay\right)\)
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
chuyển về dạng nguyên thể rồi tính thể chất khối lượng sau đó quay về đang tìm mũ của nhiều số làm ra rồi thì dễ lắm bạn ạ k minh nha
a)\(\left(x^2-2\right)\left(x^2+2x+2\right)\)
b)\(\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
c)\(-2\left(x-4\right)\left(2x+1\right)\)
d)\(\left(x-5\right)\left(4x+1\right)\)
e)\(3\left(x-2\right)\left(3x-2\right)\)
g)\(2\left(a-b\right)^2\)
h)\(\left(xy-3\right)\left(5y^2-2z\right)\)
i)\(\left(4x+1\right)\left(2x-y\right)\)
l)\(abc^2\left(b-a\right)\left(b+c\right)\)
m)\(\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Trả lời:
1) sửa đề: \(x^4+x^3-4x-4=x^3\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^3-4\right)\)
2) \(x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)=\left(x-a\right)\left(a-b\right)\)
3) \(5xy^3-2xyz-15y^2+6z=\left(5xy^3-15y^2\right)-\left(2xyz-6z\right)\)
\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)=\left(xy-3\right)\left(5y^2-2z\right)\)