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Answer:
\(25x^2-10x+4y-4y^2\)
\(=25x^2-10x+1-4x^2+4y-1\)
\(=\left(25x^2-10x+1\right)-\left(4y^2-2y+1\right)\)
\(=[\left(5x\right)^2-2.5x.1+1]-[\left(2y\right)^2-2.2y.1+1]\)
\(=\left(5x-1\right)^2-\left(2y-1\right)^2\)
\(=\left(5x-1-2y+1\right).\left(5x-1+2y-1\right)\)
\(=\left(5x-2y\right).\left(5x+2y-2\right)\)
\(25x^2-10x+1-y^2\)
\(=\left(5x-1\right)^2-y^2\)
\(=\left(5x-y-1\right)\left(5x+y-1\right)\)
\(4y^2-4x^2-4y+1\)
\(=\left(2y-1\right)^2-\left(2x\right)^2\)
\(=\left(2y+2x-1\right)\left(2y-2x-1\right)\)
\(-y^2+6y-9+x^2\)
\(=x^2-\left(y-3\right)^2\)
\(=\left(x+y-3\right)\left(x-y+3\right)\)
a) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2=\left(2x-3\right)^2\)
b) \(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
c) \(1+12x+36x^2=1^2+2.6x.1+\left(6x\right)^2=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2=\left(3x-4y\right)^2\)
f) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
g) \(-16a^4b^6-24a^5b^5-9a^6b^4=-\left(16a^4b^6+24a^5b^5+9a^6b^4\right)\)
\(=-\left[\left(4a^2b^3\right)^2+2.4a^2b^3.3a^3b^2+\left(3a^3b^2\right)^2\right]\)
\(=-\left(4a^2b^3+3a^3b^2\right)^2\)
h) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\) \(=\left(5x-2y\right)^2\)
i) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)
\(4x^3-13x^2+9x-18\)
\(=4x^3-12x^2-x^2+3x+6x-18\)
\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)
\(=\left(x-3\right)\left(4x^2-x+6\right)\)
=x2-2xy+1-4y2-4y-1
=(x-1)2-(4y2+4y+1)
=(x-1)2-(2y+1)2
=(x-1+2y+1)(x-1-2y-1)
=(x+2y)(x-2y-2)
Ta có 25x2-10xy2+y4
=(5x-y2)2 (cái này là hằng đẳng thức thứ 2 nha !!!!)
Xong rùi,nhớ
k bít làm à