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Bài 2 :
a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(A=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)
\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)
\(a,\left(x+3\right)^2\)
\(b,\left(x+\frac{1}{2}\right)^2\)
\(c,\left(xy^2+1\right)^2\)
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. $x^2-x+\dfrac14$
$=x^2-2\cdot x\cdot \dfrac12+\Bigg(\dfrac12\Bigg)^2$
$=\Bigg(x-\dfrac12\Bigg)^2$
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)\(x^2-x+\frac{1}{4}=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)