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\(a.=\left(2x\right)^2-2.2x.2y+\left(2y\right)^2=\left(2x-2y\right)^2\)
\(b.=\left(3x\right)^2-2.3x.2+2^2=\left(3x-2\right)^2\)
a. 4x2+4y2-8xy=(2x)2+(2y)2-8xy
=(2x-2y)2
b.9x2-12x+4=(3x)2-12x+22
=(3x-2)2
c.xy2+1/4x2y4+1=xy2+(1/2xy2)2+1
=(1/2xy2+2)2
a, \(25x^2+5xy+\frac{1}{4}y^2=\left(5x\right)^2+2.5x.\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)
\(=\left(5x+\frac{1}{2}y\right)^2\)
b, \(9x^2+12x+4=\left(3x\right)^2+2.3x.2+2^2=\left(3x+2\right)^2\)
c, \(x^2-6x+5-y^2-4y=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)\)
\(=\left(x-3\right)^2-\left(y+2\right)^2=\left(x-y-5\right)\left(x+y-1\right)\)
d, \(\left(2x-y\right)^2+4\left(x+y\right)^2-4\left(2x-y\right)\left(x+y\right)\)
\(=\left(2x-y\right)^2-2\left(2x-y\right)\left(2x+2y\right)+\left(2x+2y\right)^2\)
\(=\left(2x-y+2x+2y\right)^2=\left(4x+y\right)^2\)
\(4x^2-\frac{1}{9}\left(y+1\right)^2=\left(2x\right)^2-\left(\frac{1}{3}\left(y+1\right)\right)^2\)
\(=\left(2x-\frac{1}{3}\left(y+1\right)\right)\left(2x+\frac{1}{3}\left(y+1\right)\right)\)
\(=\left(2x-\frac{1}{3}y-\frac{1}{3}\right)\left(2x+\frac{1}{3}y+\frac{1}{3}\right)\)
a) Ta có: \(\left(x^2+9x+18\right)^2+2\left(x^2+9x\right)+37\)
\(=\left(x^2+9x+18\right)^2+2\cdot\left(x^2+9x+18\right)-36+37\)
\(=\left(x^2+9x+19\right)^2\)
b) Ta có: \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)
\(=\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+2\left(x+1\right)\left(y+1\right)\)
\(=\left(x^2+2x+2+y^2+2y\right)^2\)
A, \(\frac{9}{4}x^2+3x+4\)
= \(\left(\frac{3}{2}x^2\right)+2\cdot\frac{3}{2}x\cdot2+2^2\)
= \(\left(\frac{3}{2}x+2\right)^2\)
a/ 9x2-12xy+4y2 = (3x - 2y)2
b/ 25x2-10x+1 = (5x - 1)2
c/ 9x2-12x+4 = (3x - 2)2
d/ 4x2+20x+25 = (2x + 5)2
e/ x4-4x2+4 = (x2 - 2)2
\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)
Học tốt <3
x(3x-1)+(9x-5)(x-2)=3x2-x+9x(x-2)-5(x-2)=3x2-x+9x2-18x-5x+10=12x2-22x+10
a)4x2+12xy+9y2= (2x)2+2.2x.3y+(3y)2=(2x+3y)2
b)y2+1-2y= y2-2.y.1+12=(y-1)2
\(\left(9x^2-12x+4\right)-\left(y+2\right)^2\)
\(=\left[\left(3x^2\right)-2.3x.2+2^2\right]-\left(y+2\right)^2\)
\(=\left(3x-2\right)^2-\left(y+2\right)^2\)
\(\left(9x^2-12x+4\right)-\left(y+2\right)^2\)
= \(9x^2-12x+4-\left(y^2+4y+4\right)\)
=\(9x^2-12x+4-y^2-4y-4\)
=\(9x^2-y^2-12x-4y\)
=\(\left(3x-y\right)\left(3x+y\right)-4\left(3x+y\right)\)
=\(\left(3x+y\right)\left(3x-y-4\right)\)