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a, Ta có x2 - x - y2 - y
= ( x2 - y2 ) - ( x + y )
= ( x - y ).( x + y ) - ( x + y )
= ( x+ y ).( x - y -1 )
b, Ta có x2 - 2xy + y2 - z2
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - z2
= ( x - y - z ).( x - y + z )
a) x2 - x - y2 - y = x2 - y2 - x - y
=(x - y) (x + y) - (x + y)
=(x + y) (x - y - 1)
b) x2 - 2xy + y2 - z2 = (x - y)2 - z2
=(x - y- z) (x - y + z)
a) x\(^2\)-x-y\(^2\)-y
=(x\(^2\)-y\(^2\)) - (x-y)
=xy(x-y) - (x-y)
=xy(x-y)
Ta có:
\(x^2+2xy+y^2-x-y-12=(x^2+2xy+y^2)-(x+y)-12\)
\(=(x+y)^2-(x+y)-12 \) \((*)\)
Đặt \(x+y=a\)
từ \((*)\Rightarrow a^2-a-12=(a^2+3a)-(4a+12)\)
\(=(a+3)(a-4)\)
Thay \(a=x+y\)
\(\Rightarrow (x+y+3)(x+y-4)\)
16-2xy-x2-y2
= - (x2+2xy+y2-16)
=- ( ( x+y)2- 42 )
làm tiếp nha bạn !
a) x2+ 4x+4-y2
=(x2+2.x.2+22)-y2
=(x+2)2-y2
=(x+2+y)(x+2-y)
b)(x2-2xy+y2)-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
\(x^2+4x+4-y^2\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
hk tốt
^^
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
\(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3.\left[\left(x+y\right)^2-z^2\right]=3.\left(x+y-z\right)\left(x+y+z\right)\)
\(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
\(=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y+2xy\right)\left(x+y-2xy\right)\)
x2 -4x2y2 +y2 +2xy =( x2 +2xy +y2) -(2xy)2 =(x+y)2 -(2xy)2 =(x+y+2xy)(x+y-2xy)
\(x^2-2xy+y^2-z^2\)
Áp dụng hằng đẳng thức:\(\left(a-b\right)^2=a^2-2ab+b^2\)
\(=\left(x-y\right)^2-z^2\)
Áp dụng hằng đẳng thức:\(a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)