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

b) \(x^2+x=\text{ }\left[x^2+2.\frac{1}{2}x+\left(\frac{1}{2}\right)^2\right]-\left(\frac{1}{2}\right)^2=\left(x+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\) 

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

a) \(9x^2+6x+1=\left(3x+1\right)^2\)

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

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

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

a) 9x² + 6x + 1 = ( 3x + 1 )²

b) x² - x + 1/4 = ( x - 1/2)²

c) x² . y⁴ - 2xy² + 1 = ( xy² - 1 ) ²

d) x² + 2/3x + 1/9 = (x+1/3)²

a) \(9x^2+6x+1\)

\(=\left(3x\right)^2+6x+1\)

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

b) \(x^2-x+\frac{1}{4}\)

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

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

c)  \(x^2y^4-2xy^2+1\)

.... Rút gọn lũy thừa cao

d ) Tương tự câu b . Làm theo cách thêm bớt

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) \(x^2+6x+9=\left(x+3\right)^2\)

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.

Bài 8:

Ta có: \(A=-x^2+2x+4\)

\(=-\left(x^2-2x-4\right)\)

\(=-\left(x^2-2x+1-5\right)\)

\(=-\left(x-1\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=1

 lớp 7 đã hok hằng đẳng thức đâu : nhưng thôi đây vẫn giúp em

x² + 6x + 9 

= x² + 2 .x . 3 + 3² 

= ( x + 3 )² 

x² + 5x + 25 / 4

= x² + 2 . x . 5/ 2 + ( 5 / 2 )² 

= ( x + 5/2 )² 

x² + x + 1 /4 

= x² + 2 . x . 1/2 + ( 1/2 ) ² 

= ( x + 1/2 )² 

study well 

2xy² +x² + y⁴ + 1 

= x² + 2xy² + y⁴ + 1 

= ( x + y² )² + 1 

Study well 

chúc e hok tốt 

Hình như bn bỏ nhầm lớp

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

2)\(x^2+5x+\frac{25}{4}=x^2+2x.\frac{5}{2}+\left(\frac{5}{2}\right)^2=\left(x+\frac{5}{2}\right)^2\)

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

4)Bn ghi mik k hiểu cho lắm nên k lm nha