Bài 1: 

a) \(a^2-6a+9=\left(a-3\right)^2\)

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

Bài 2:

a)  \(\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\)

\(\Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\)

b) \(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

này mình có vài câu không làm được, xin lỗi bạn nha

\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)

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

                                       \(=\left(2x-\frac{1}{3}\left(y+1\right)\right)\left(2x+\frac{1}{3}\left(y+1\right)\right)\)

                                       \(=\left(2x-\frac{1}{3}y-\frac{1}{3}\right)\left(2x+\frac{1}{3}y+\frac{1}{3}\right)\)

`a,-x^3/8 + 3/(4x^2) - 3/(2x) +1`

`=-(x^3/8 - 3/(4x^2) + 3/(2x) - 1)`

`=-(x/2 - 1)^3`

`b,x^6 - 3/(2x^{4} y) + 3/(4x^{2}y^{2}) - 1/(8y^{3})`

`=(x^3 - 1/(2y))^{3}`

Bài làm:

Ta có: \(\frac{9}{4x^2}+\frac{9y^2}{4}-\frac{9y}{2x}\)

\(=\left(\frac{3}{2x}\right)^2-2.\frac{3}{2x}.\frac{3y}{2}+\left(\frac{3y}{2}\right)^2\)

\(=\left(\frac{3}{2x}-\frac{3y}{2}\right)^2\)

dạ mk cảm ơn bạn De Bruyne nha!

`a, a^2 + 10ab + 25b^2 = (a+5b)^2`

`b, 1 + 9a^2 - 6a = (3a-1)^2`

a) \(a^2+10ab+25b^2\)

\(=a^2+2\cdot5b\cdot a+\left(5b\right)^2\)

\(=\left(a+5b\right)^2\)

b) \(1+9a^2-6a\)

\(=1-6a+9a^2\)

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

a) a² + 10ab + 25b²

= a² + 2.a.5b + (5b)²

= (a + 5b)²

b) 1 + 9a² - 6a

= 9a² - 6a + 1

= (3a)² - 2.3a.1 + 1²

= (3a - 1)²

a: \(25x^2-\dfrac{10}{3}xy+\dfrac{1}{9}y^2=\left(5x-\dfrac{1}{3}y\right)^2\)

b: \(25x^2-15x+\dfrac{9}{4}=\left(5x-\dfrac{3}{2}\right)^2\)

c: \(\left(2x+\dfrac{1}{2}y\right)\left(4x^2-xy+\dfrac{1}{4}y^2\right)=8x^3+\dfrac{1}{8}y^3\)

d: \(\left(x^2-\dfrac{2}{3}\right)\left(x^4+\dfrac{2}{3}x^2+\dfrac{4}{9}\right)=x^6-\dfrac{8}{27}\)

a) \(x^2+2x+1=x^2+2\cdot x\cdot1+1^2=\left(x+1\right)^2\)

b) \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)

c) \(x^2+6xy+9y^2=x^2+2\cdot x\cdot3y+\left(3y\right)^2=\left(x+3y\right)^2\)

d) \(z^2-z+\dfrac{1}{4}=z^2-2\cdot z\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(z-\dfrac{1}{2}\right)^2\)

e) \(25x^2-10x+1=\left(5x\right)^2-2\cdot5x\cdot1+1^2=\left(5x-1\right)^2\)

\(a,x^2+5x+\frac{25}{4}\)

\(=x^2+2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2\)

\(=\left(x+\frac{5}{2}\right)^2\)