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

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

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

a)x²-6x+9

=x²-2.x.3+3²

=(x-3)²

b)4x²+4x+1

=(2x)²+2.2x.1+1²

=(2x+1)²

c)4x²+12xy+9y²

=(2x)²+2.2x.3y+(3y)²

=(2x+3y)²

d)4x⁴-4x²+4

=(2x²)²-2.2x².2+2²

=(2x²-2)²

a: \(\left(3x-1\right)\left(9x^2+3x+1\right)=27x^3-1\)

b: \(\left(1-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{5}+1\right)=1-\dfrac{x^3}{125}\)

c: \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)=x^3+27y^3\)

d: \(\left(4x+3y\right)\left(16x^2-12xy+9y^2\right)=64x^3+27y^3\)

a) \(x^2+4x+4\)

\(=x^2+2\cdot2\cdot x+2^2\)

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

b) \(4x^2-4x+1\)

\(=\left(2x\right)^2-2\cdot2x\cdot1+1^2\)

\(=\left(2x-1\right)^2\)

c) \(x^2-x+\dfrac{1}{4}\)

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

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

d) \(4\left(x+y\right)^2-4\left(x+y\right)+1\)

\(=\left[2\left(x+y\right)\right]^2-2\cdot2\left(x+y\right)\cdot1+1^2\)

\(=\left[2\left(x+y\right)-1\right]^2\)

\(=\left(2x+2y-1\right)^2\)

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

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

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

a) ( x   +   1 ) 2 .                     b) ( x   –   4 ) 2 .

c) x 2 4 + x + 1 ;                   d) ( 2 x   –   2 y ) 2 .

a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

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

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

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

c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

\(=\left(x-y-2\right)^2\)

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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

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