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9x2+4y2+2(3x+2y+6xy)+1
= 9x2+4y2+1+6x+4y+12xy
=(3x)2+(2y)2+12+2.3x.2y+2.2y.1+2.3x.1 (1)
Thay 3x=m,2y=n,1=p
=>(1)=m2+n2+p2+2mn+2np+2pm=(m+n+p)2
=> 9x2+4y2+2(3x+2y+6xy)+1=(3x+2y+1)2
a) = x^2 - 2x + 1 + 4y^2 + 4y + 1
= ( x - 1 )^2 + ( 2y + 1 )^2
b) = 4x^2 + 4x +1 + 4y^2 + 4y + 1
= ( 2x + 1 )^2 + ( 2y + 1 )^2
c) = 9x^2 - 12x + 4 + 16y^2 - 24y + 9
=( 3x - 2 )^2 + ( 4y - 3 )^2
d) = 4x^2 + 4xy+ y^2 + x^2 - 2xz + z^2
= ( 2x + y )^2 + ( x - z )^2
Lời giải:
Ta có:
\((2x-4y)^2+4x-8y+1=(2x-4y)^2+2(2x-4y)+1^2\)
\(=(2x-4y+1)^2\)
a) \(9x^2+6x+1=\left(3x+1\right)^2\)
b)\(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)
c)\(x^2y^4-2xy^2+1=\left(xy^2-1\right)^2\)
d) \(x^2+\frac{2}{3}x+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2\)
a) 9x2 + 6x + 1 = ( 3x + 1 )2
b) x2 - x + 1/4 = ( x - 1/2)2
c) x2 . y4 - 2xy2 + 1 = ( xy2 - 1 ) 2
d) x2 + 2/3x + 1/9 = (x+1/3)2
(2x+3y)2+2(2x+3y)+1=[(2x+3y)+1)]2=(2x+3y+1)2
9x2-6x+1=(3x)2-2.3x+12=(3x-1)2
=(3x-1)^2+(2y-2)^2-5
= ( 3x-1 )^2+(2y-2)^2-5