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x^2-y^2+10x-6y-9
=x^2+10x -(y^2+6y+9)
= x^2+10x-(y^2+2.3.y+3^2)
=x^2+10x-(y+3)^2
=\(\left[x^2-\left(y+3\right)^2\right]+10x \)
={\(\left(x+x+3\right)\left[x-\left(x+3\right)\right]\)}\(+10x\)
=\(\left(x+x+3\right).\left(x-x-3\right)+10x\)
(đến đây b tự lm nhé, m cx k bt đúng k nx )


\(x^2-y^2+10x-6y+16\)
\(=\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)
\(=\left(x+5\right)^2-\left(y+3\right)^2\)
\(=\left(x+5-y-3\right)\left(x+5+y+3\right)\)
\(=\left(x-y+2\right)\left(x+y+8\right)\)

1) \(x^2+6y-9-y^2=x^2-\left(y^2-6y+9\right)=x^2-\left(y-3\right)^2=\left(x-y+3\right)\left(x+y-3\right)\)
2) \(9y^2-6y+1-25x^2=\left(3y\right)^2-2.3y+1-\left(5x\right)^2=\left(3y-1\right)^2-\left(5x\right)^2\)
\(=\left(3y-1-5x\right)\left(3y-1+5x\right)\)
3) \(a^2-9+6x-x^2=a^2-\left(x^2-6x+9\right)=a^2-\left(x-3\right)^2=\left(a-x+3\right)\left(a+x-3\right)\)


\(x^2-y^2+6y-9=x^2-\left(y^2-6y+9\right)\)
\(=x^2-\left(y-3\right)^2\)
\(=\left(x+y-3\right)\left(x-y+3\right)\)

bài 1:= \(2x\left(x-3\right)-6\left(x-3\right)+2y\left(x-3\right)\)
=\(2\left(x-3\right)\left(x+y-3\right)\)
bài 2:P=\(x^2-2x+1+y^2+6y+9+2\)
P=\(\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)
vậy Pmin=2 khi x=1 và y=-3
Ta có:x^2-y^2-6y-9=x^2-(y^2+6y+9)
=x^2-(y+3)^2
=(x-y-3)(x-y+3)
Ta có :
x2 - y2 + 6y - 9
= x2 - ( y2 + 6y + 9 )
= x2 - ( y + 3 )2
= ( x - y - 3 ) ( x - y + 3 )