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Ta có
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) 7xy^2+5x^2y
= xy(7y+5x)
b) x^2+2xy+y^2-11x-11y
= (x^2+2xy+y^2)-(11x+11y)
=(x+y)^2-11(x+y)
=(x+y)(x+y-11)
a) 7xy2 + 5x2y
= xy ( 7y + 5x )
b) x2 + 2xy + y2 - 11x - 11y
= ( x2 + 2xy + y2 ) - ( 11x + 11y )
= ( x + y )2 - 11 ( x + y )
= ( x + y )( x + y - 11 )
\(=3^2-\left(x-y\right)^2=\left[3-\left(x-y\right)\right]\left[3+\left(x-y\right)\right]=\left(3-x+y\right)\left(3+x-y\right)\)
\(9-x^2+2xy-y^2\)
\(=9-\left(x^2-2xy+y^2\right)\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3-x+y\right)\left(3-x-y\right)\)
a, = (xy-y^2) + (2x-2y) = y(x-y) + 2.(x-y) = (x-y).(y+2)
b, = (x+y)^2 - 9 = (x+y-3).(x+y+3)
\(\left(a^2+b^2-c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2\right)^2-\left(2ab\right)^2\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[\left(a-b\right)^2+c^2\right]\)
=(a+b+c)(a+b-c)(a-b+c)(a-b-c)
\(a,=\left(x+y\right)^2-5^2=\left(x+y+5\right)\left(x+y-5\right)\)
\(b,=\left(a-1\right)^2-1-\left(2b+1\right)^2-1=\)
\(c,=\left(a-b\right)\left(a+b\right)-5\left(a+b\right)=\left(a+b\right)\left(a-b-5\right)\)
a,( x2 + 2xy + y2 ) -25
=( x + y )2 - 52
=( x + y + 5) ( x + y - 5)
b,
a) \(a^5+a^3-a^2-1\)
\(=a^5+a^4+a^3+a^3+a^2+a-a^4-a^3-a^2-a^2-a-1\)
\(=a^3\left(a^2+a+1\right)+a\left(a^2+a+1\right)-a^2\left(a^2+a+1\right)-\left(a^2+a+1\right)\)
\(=\left(a^3+a-a^2-1\right)\left(a^2+a+1\right)\)
\(=\left[\left(a^3-1\right)-a\left(a-1\right)\right]\left(a^2+a+1\right)\)
\(=\left[\left(a-1\right)\left(a^2+a+1\right)-a\left(a-1\right)\right]\left(a^2+a+1\right)\)
\(=\left(a-1\right)\left(a^2+a+1-a\right)\left(a^2+a+1\right)\)
\(=\left(a-1\right)\left(a^2+1\right)\left(a^2+a+1\right)\)
b) \(27a^2b^2-18ab+3\)
\(=3\left(9a^2b^2-6ab+1\right)\)
\(=3\left(3ab-1\right)^2\)
c) \(4-x^2-2xy-y^2\)
\(=4-\left(x+y\right)^2\)
\(=\left(2-x-y\right)\left(2+x+y\right)\)
Cam on nhe