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

a)\(x^2+2x+1=x^2+2x1+1^2=\left(x+1\right)^2\)

b)\(9x^2+y^2+6xy=3^2x^2+y^2+2.3x.y=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)

c)\(25a^2+4b^2-20ab=5^2a^2+2^2b^2-2.5a.2b=\left(5a\right)^2-2.5a.2b+\left(2b\right)^2=\left(5a-2b\right)^2\)

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

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

\(\frac{1}{4}a^2+2ab+4b^2=\left(\frac{1}{2}a+b\right)^2\)

\(25+10x+x^2=\left(x+5\right)^2\)

\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(y^4-\frac{1}{3}\right)^2\)

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.

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, \(25x^2+5xy+\frac{1}{4}y^2=\left(5x\right)^2+2.5x.\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)

\(=\left(5x+\frac{1}{2}y\right)^2\)

b, \(9x^2+12x+4=\left(3x\right)^2+2.3x.2+2^2=\left(3x+2\right)^2\)

c, \(x^2-6x+5-y^2-4y=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)\)

\(=\left(x-3\right)^2-\left(y+2\right)^2=\left(x-y-5\right)\left(x+y-1\right)\)

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

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

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

Có cái cc

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) x²+x+\(\dfrac{1}{4}\)=x²+2.x.\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)=(x+\(\dfrac{1}{2}\))²

c) 2xy²+x²y⁴+1=(xy²)²+2.xy²+1=(xy²+1)²

a,(x+3)^2

b,(x+1/2)^2

a) (x+3)²

b) (x+\(\dfrac{1}{2}\))²

c) (xy²+1)²