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\(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\left(\frac{y-x}{xy}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\frac{\left(x-y\right)^2}{x^2y^2}-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2x^2y^2}{xy\left(x-y\right)^2}-\frac{x^2+y^2}{\left(x-y\right)^2}\)
\(=\frac{2xy}{\left(x-y\right)^2}-\frac{x^2+y^2}{\left(x-y\right)^2}=\frac{-x^2+2xy-y^2}{\left(x-y\right)^2}\)
\(=-\frac{\left(x-y\right)^2}{\left(x-y\right)^2}=-1\)
\(\frac{12}{5}x^2y^2-9x^4-\frac{4}{25}y^4\)
\(=-\left(\frac{4}{25}y^4-\frac{12}{5}x^2y^2+9x^4\right)\)
\(=-\left[\left(\frac{2}{5}y^2\right)^2-2\cdot\frac{2}{5}y^2\cdot3x^2+\left(3x^2\right)^2\right]\)
\(=-\left(\frac{2}{5}y^2-3x^2\right)^2\)
\(\frac{12}{5}x^2y^2-9x^4-\frac{4}{25}y^4\)
\(=-\left(9x^4-\frac{12}{5}x^2y^2+\frac{4}{25}y^4\right)\)
\(=-\left[\left(3x\right)^2-2.3x^2.\frac{2}{5}y^2+\left(\frac{2}{5}y^2\right)^2\right]\)
\(=-\left(3x^2+\frac{2}{5}y^2\right)\)
x^3-5x2+8x-4=x3-2x2-3x2+6x+2x-4=x2(x-2)-3x(x-2)+2(x-2)
=(x-2)(x2-3x+2)
=(x-2)(x2-2x-x+2)=(x-2)(x-2)(x-1)=(x-2)2(x-1)
x2-2.x.1/2 +(1/2)2-9/4
=(x-1/2)2-9/4
=(x-1/2)2-(3/2)2
=(x-1/2-3/2).(x-1/2+3/2)
=(x-2)(x+1)
Ta có : \(F=x^2-4^x+4-y^2\)
\(=\left(x^2-4^x+4\right)-y^2\)( nhóm hạng tử )
\(=\left(x-2\right)^2-y^2\)( đẳng thức số 2 )
\(=\left(x-2-y\right)\left(x-2+y\right)\)( đẳng thức số 3 )
Vậy : \(F=\left(x-2-y\right)\left(x-2+y\right)\)
x2 + 1 - y2 - 2x
= x2 - 2x + 1 - y2
=[x2 - 2x + 1] - y2
=[x-1]2 - y2
=[x-1-y][x-1+y]
a) \(x^2+1-y^2-2x=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
b) \(64x^4+y^4=\left(8x^2\right)^2+\left(y^2\right)^2=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
x4 + 2x3 + x2 - y2
= ( x4 + 2x3 + x2 ) - y2
= [ ( x2 )2 + 2.x2.x + x2 ] - y2
= ( x2 + x )2 - y2
= ( x2 + x - y )( x2 + x + y )
\(=x^2\left(x^2+2x+1\right)-y^2\)
\(=x^2\left(x+1\right)^2-y^2\)
\(=x^2\left(x+1-y\right)\left(x+1+y\right)\)