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sai rồi vì (x2+6xy+9) ko có y2 nên ko thể có hằng đẳng thức được
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a) 9x^2+6xy+y^2
=(3x+y)2
b)6x - 9 - x^2
=-(x2-6x+9)
=-(x-3)2
c) x^2 + 4y^2 + 4xy
=(x+2y)2
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a. \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)
b. \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)
a: \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)
b: \(x^2-6xy+9y^2-36=\left(x-3y\right)^2-6^2=\left(x-3y-6\right)\left(x-3y+6\right)\)
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a) Ta có: \(x^2y^2-x^2+6xy-9y^2\)
\(=x^2y^2-\left(x^2-6xy+y^2\right)\)
\(=\left(xy\right)^2-\left(x-3y\right)^2\)
\(=\left(xy-x+3y\right)\left(xy+x-3y\right)\)
b) Ta có: \(9-x^2+2xy-y^2\)
\(=9-\left(x^2-2xy+y^2\right)\)
\(=9-\left(x-y\right)^2\)
\(=\left(9-x+y\right)\left(9+x-y\right)\)
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9x2 + 6xy + y2
= (3x)2 + 2.3x.y + y2
= (3x + y)2
b) 6x - 9 - x2
= -(x2 - 6x + 9)
= -(x - 3)2
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\(5x^2-6xy+y^2=\left(9x^2-6xy+y^2\right)-4x^2=\left(3x-y\right)^2-4x^2=\left(3x-y-2x\right)\left(3x-y+2x\right)=\left(x-y\right)\left(5x-y\right)\)
\(5x^2-6xy+y^2\)
\(=5x^2-5xy-xy+y^2\)
\(=5x\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)\left(5x-y\right)\)
9 x 2 + 6xy + y 2 = 3 x 2 + 2.(3x)y + y 2 = 3 x + y 2