a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)

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

b) \(x^2+y^2+2xy=\left(x+y\right)^2\)

c) \(9x^2+12x+4=\left(3x+2\right)^2\)

d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)

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

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

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

A, \(\frac{9}{4}x^2+3x+4\)

= \(\left(\frac{3}{2}x^2\right)+2\cdot\frac{3}{2}x\cdot2+2^2\)

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

B. ( 9\(x^2\)+ 12x+ 4) + 6(3x+2)+9

=[ (3x)\(^2\)+2.3x.2+\(2^2\) ) + 2.(3x+2).3 +3\(^2\)

= ( 3x+2)\(^2\)+2.(3x+2).3+3\(^2\)

= (3x+2+3)\(^2\)

Chúc bạn học tốt <3

trả lời 

=(3/x+2)²

chúc bn 

học tốt

a) ( 2 x   +   1 ) 2 .                     b) ( 3 x   –   2 ) 2 .

c) 1 2 ab 2 + 1 2 .                  d) ( 4 uv ²   –   1 ) 2 .

b. (9x² +12x + 4) + 6(3x + 2) + 9

= (3x+2)^2 + 2.3.(3x+2) + 3^2

= (3x+2+3)^2

= (3x+5)^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, \(8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3=\left(2x+y\right)^3\)

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`

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

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