a) \(9x^2+6x+1=\left(3x\right)^2+2.3x.1+1^2=\left(3x+1\right)^2\)

b) \(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2=\left(x-0,5\right)^2\)

c) \(x^2y^4-2xy^2+1=\left(xy^2\right)^2-2.xy^2.1+1^2=\left(xy^2-1\right)^2\)

d) \(x^2+\dfrac{2}{3}x+\dfrac{1}{9}=x^2+2.x.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2=\left(x+\dfrac{1}{3}\right)^2\)

a) \(9x^2+6x+1\)

\(=\left(3x\right)^2+2.3x.1+1^2\)

\(=\left(3x+1\right)^2\)

b) \(x^2-x+\dfrac{1}{4}\)

\(=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)

\(=\left(x-\dfrac{1}{2}\right)^2\)

\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)

Bài 1 em dùng HĐT nha

Bài 2:

a. x² + 4x + 4

= x² + 2.2.x + 2²

= (x + 2)²

b. 9x² - 12x + 4

= (3x)² - 3x.2.2 + 2²

= (3x - 2)²

c. \(\dfrac{x^2}{4}+x+1\)

= \(\left(\dfrac{x}{2}\right)^2+2.\dfrac{x}{2}.1+1^2\)

= \(\left(\dfrac{x}{2}+1\right)^2\) 

a) \(x^2+6x+9=\left(x+3\right)^2\)

b) \(2xy^2+x^2y^4+1=x^2y^4+2xy^2+1=\left(xy^2+1\right)^2\)

c) \(x^2+x+\frac{1}{4}=x^2.2.\frac{1}{2}x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2\)

hay thật để 1 tiếng mà chả ai làm được.

a=(x+3)

b=(x+1/2)

c=(xy^2+1)

Good luck!

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

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

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

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

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

\(\frac{1}{4}a^2+2ab+4b^2=\left(\frac{1}{2}a+b\right)^2\)

\(25+10x+x^2=\left(x+5\right)^2\)

\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(y^4-\frac{1}{3}\right)^2\)

a) x²+6x+9=x²+2.x.3+3²=(x+3)²

b) x²+x+\(\dfrac{1}{4}\)=x²+2.x.\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)=(x+\(\dfrac{1}{2}\))²

c) 2xy²+x²y⁴+1=(xy²)²+2.xy²+1=(xy²+1)²

a,(x+3)^2

b,(x+1/2)^2

a) (x+3)²

b) (x+\(\dfrac{1}{2}\))²

c) (xy²+1)²

a) \(6x^2y+9+x^4y^2=\left(x^2y+3\right)^2\)

b) \(-4xy+4x^2+y^2=\left(2x-y\right)^2\)

c) \(25y^4-10y^2+1=\left(5y^2-1\right)^2\)

\(a,=\left(x^2y+3\right)^2\\ b,=\left(2x+y\right)^2\\ c,=\left(5y^2-1\right)^2\)