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a)\(\left(x-y\right)^2-2\left(x-y\right)+1=\left(x-y-1\right)^2\)
b)\(x^2-2y-1-2x+1-y^2=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left[\left(x-1\right)-\left(y+1\right)\right]\left[\left(x-1\right)+\left(y+1\right)\right]\)
\(=\left(x-y-2\right)\left(x+y\right)\)
c)\(x^2-y^2-2x-1=x^2-\left(y^2+2x+1\right)\)
\(=x^2-\left(y+1\right)^2\)
\(=\left(x^2-y-1\right)\left(x^2+y+1\right)\)
A. Ta có: (x - y)2 - 2(x - y)+1 = (x - y)2 - 2.(x - y).1 +12 = ( x - y - 1)2
B. Ta có: x2 - 2y -1 - 2x +1 -y2 = (x2 - y2) - (2x - 2y) -1+1 = (x - y)(x + y) - 2(x - y) = (x - y)(x + y - 2)
C. Ta có: x2 - y2 -2y -1 = x2 -(y2 - 2y -1) = x2 - ( y2 +2y1 + 1) = x2 - (y+1)2 = (x - y - 1)(x + y +1)
k cho mình nha bạn hihj!!! ~3~
a/ \(12x^2+5x-12y^2+12y-10xy-3.\)
\(=12x^2+9x-4x-12y^2+6y+6y-18xy+8xy-3.\)
\(=\left(12x^2-18xy+9x\right)-\left(4x-6y+3\right)+\left(8xy-12y^2+6y\right)\)
\(=3x\left(4x-6y+3\right)-\left(4x-6y+3\right)+2y\left(4x-6y+3\right)\)
\(=\left(4x-6y+3\right)\left(3x-1+2y\right)\)
2/ \(2x^2+y^2+3x-2y-3xy+1\)
\(=\left(y^2-2y+1\right)+\left(3x-3xy\right)+2x^2\)
\(=\left(y-1\right)^2+3x\left(1-y\right)+2x^2\)
\(=\left(y-1\right)^2-3x\left(y-1\right)+2x^2\)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
8x3 - 27y3 = 23 . x3 - 33 . y3 = ( 2x )3 - ( 3y )3 = ( 2x - 3y ) [(2x)2 + 12xy + (3y)2 ].
\(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)
\(\Leftrightarrow\left(1+x^2\right)^2+4x\left(1+x^2\right)\)
\(\Leftrightarrow\left(1+x^2\right)\times\left[\left(1+x^2\right)+4\right]\)
( 1+x2 )2 -4x( 1- x2 )
=x4+2x2+1-4x+4x3
=x3+2x2-x+2x3+4x2-2x-x2-2x+1
=x(x2+2x-1)+2x(x2+2x-1)-(x2+2x-1)
=(x2+2x-1)(x2+2x-1)
=(x2+2x-1)2
\(\left(1+x\right)^2-4x\left(1-x^2\right)\)
\(=\left(1+x\right)^2-4x\left(1-x\right)\left(1+x\right)\)
\(=\left(1+x\right)\left(1+x-4\left(1-x\right)\right)\)
\(=\left(1+x\right)\left(1+x-4+4x\right)\)
\(=\left(1+x\right)\left(5x-3\right)\)
\(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(\left(xy+1-x-y\right)\left(xy+1+x+y\right)\)
\(=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)