Bài 2 :

a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)

\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)

\(A=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)

\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)

a=(x+3)

b=(x+1/2)

c=(xy^2+1)

Good luck!

\(a,\left(x+3\right)^2\)

\(b,\left(x+\frac{1}{2}\right)^2\)

\(c,\left(xy^2+1\right)^2\)

Ta có: (a+b+c)^2 + a^2 + b^2 + c^2

= a^2 +b^2 +c^2 + 2ab + 2ac + 2bc + a^2 + b^2 + c^2

= (a^2 +2ab+ b^2) + (b^2 +2bc+ c^2) +(c^2 +2ac+ a^2 )

= (a+b)^2 +(b+c)^2 +(c+a)^2

\(\left(a^2+b^2+c^2\right)+a^2+b^2+c^2\)

\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)

a) \(x^2+6x+9=x^2+2.3x+3^2=\left(x+3\right)^2\)

b) \(x^2+x=\text{ }\left[x^2+2.\frac{1}{2}x+\left(\frac{1}{2}\right)^2\right]-\left(\frac{1}{2}\right)^2=\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\) 

c) \(2xy^2+x^2y^4=\left[\left(xy^2\right)^2+2.xy^2+1^2\right]-1^2=\left(xy^2+1\right)^2-1^2\)

a. $x^2+4x+4$

$=x^2+2\cdot x\cdot2+2^2$

$=(x+2)^2$

b. $x^2-6xy+9y^2$

$=x^2-2\cdot x\cdot3y+(3y)^2$

$=(x-3y)^2$

c. $4x^2+12x+9$

$=(2x)^2+2\cdot2x\cdot3+3^2$

$=(2x+3)^2$

d. $x^2-x+\dfrac14$

$=x^2-2\cdot x\cdot \dfrac12+\Bigg(\dfrac12\Bigg)^2$

$=\Bigg(x-\dfrac12\Bigg)^2$

`x^2 +4x+4`

`=x^2+2*x*2+2^2`

`=(x+2)^2`

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`x^2-6xy+9y^2`

`=x^2 - 2*x*3y+(3y)^2`

`=(x-3y)^2`

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`4x^2 +12x+9`

`=(2x)^2 +2*2x*3+3^2`

`=(2x+3)^2`

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`x^2-x+1/4`

`=x^2 - 2*x*1/2 +(1/2)^2`

`=(x+1/2)^2`

a)\(x^2+2x+1=x^2+2x1+1^2=\left(x+1\right)^2\)

b)\(9x^2+y^2+6xy=3^2x^2+y^2+2.3x.y=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

c)\(25a^2+4b^2-20ab=5^2a^2+2^2b^2-2.5a.2b=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2=\left(5a-2b\right)^2\)

d)\(x^2-x+\frac{1}{4}=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)