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a,\(\left(x-5\right)^2-16=\left(x-5\right)^2-4^2=\left(x-5-4\right)\left(x-5+4\right)=\left(x-9\right)\left(x-1\right)\)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
a) \(x^3+3.2x^2y+3.2^2.x.y^2+\left(2y\right)^3=\left(x+2y\right)^3\)
b) áp dụng HDT : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Rightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=3x\left(x+2\right)\)
c) cũng áp dụng hdt :\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2=\left[3\left(x+5\right)-x+7\right]\left[3\left(x+5\right)+x-7\right]\)\(=\left(3x+15-x+7\right)\left(2x+15+x-7\right)=\left(2x+22\right)\left(3x+8\right)=2\left(x+11\right)\left(3x+8\right)\)
d) áp dụng típ \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)=\left(x-9y\right)\left(9x-y\right)\)
e)Áp dụng típ Hdt như trên
\(\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2=\left[7\left(y-4\right)-3\left(y+2\right)\right]\left[7\left(y-4\right)+3\left(y+2\right)\right]\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)=\left(4y-34\right)\left(11y-22\right)\)
\(=2\left(2y-17\right).11\left(y-2\right)=22\left(2y-17\right)\left(y-2\right)\)
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