1, \(x^2+2xy+y^2=\left(x+y\right)^2\)

2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)

3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)

4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)

5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

1: =(x+y)^2

2: =(2x+3)^2

3: =(x+5/2)^2

4: =(4x-1)^2

5: =(x+1/2)^2

6: =(x-3/2)^2

7: =(x+1)^3

8: =(1/2x+1)^2

9: =(3y-1/3)^3

10: =(2x+y)^3

6, \(x^2-3x+\dfrac{9}{4}=x^2-2\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2=\left(x-\dfrac{3}{2}\right)^2\)

7, \(x^3+3x^2+3x+1=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x+1\right)^3\)

8, \(\dfrac{x^2}{4}+x+1=\left(\dfrac{x}{2}\right)^2+2\cdot\dfrac{x}{2}\cdot1+1^2=\left(\dfrac{x}{2}+1\right)^2\)

9, \(27y^3-9y^2+y-\dfrac{1}{27}=\left(3y\right)^3-3\cdot\left(3y\right)^2\cdot\dfrac{1}{3}+3\cdot3y\cdot\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3=\left(3y-\dfrac{1}{3}\right)^3\)

10, \(8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3=\left(2x+y\right)^3\)

Cậu check lại đề đi xem có nhầm ở đâu không ?

`6x` hay `3x` cậu nhỉ nếu `6x thì :

`9/4x^2+6x+4`

`=(3/2x)^2 + 2. 3/2x .2 +2^2`

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

\(a,=\left(x+\dfrac{5}{2}\right)^2\\ b,=\left(2x+3y\right)^2\\ c,=a^2+b^2+c^2+2ab-2bc-2ac\\ d,=\left(4x-1\right)^2\\ e,=a^2+b^2+c^2+2ab+2bc+2ac\\ f,=a^2+b^2+c^2-2ab+2bc-2ac\)

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

c: \(x^6-3x^5+3x^4-x^3=\left(x^2-x\right)^3\)

\(b,=\left(x^4y^8\right)^2+2\cdot2x^4y^8+2^2=\left(x^4y^8+2\right)^2\)

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!

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

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

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