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

x 2  + x + 1/4 =  x 2  + 2.x.1/2 +  1 / 2 2  =  x + 1 / 2 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) \(x^2+4x+4\)

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

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

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

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

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

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

=x^2+2.x.1/2+1/4= (x+1/2)^2

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

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

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

x^2+x+1/4

=(x)^2+2.x.1/2+(1/2)^2

=(x+1/2)

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

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

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

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

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

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

a. (x + y)² = x² + 2xy + y²

b. (x - 2y)² = x² - 4xy - 4x²

c. (xy² + 1)(xy² - 1) = x²y⁴ - 1

d. (x + y)²(x - y)² = (x² + 2xy + y²)(x² - 2xy + y²) = x⁴ - (2xy + y²)² = x⁴ - (4x²y² + y⁴) = x⁴ - 4x²y² - y⁴

^{Chucs hocj toots}

Câu 2: 

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(x^2+10x+25=\left(x+5\right)^2\)

d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)

e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)

Câu 7:

a: Ta có: \(A=x^2-2x+7\)

\(=x^2-2x+1+6\)

\(=\left(x-1\right)^2+6\ge6\forall x\)

Dấu '=' xảy ra khi x=1

b: Ta có: \(B=5x^2-20x\)

\(=5\left(x^2-4x+4-4\right)\)

\(=5\left(x-2\right)^2-20\ge-20\forall x\)

Dấu '=' xảy ra khi x=2

\(x^2+2\left(x+1\right)^2+3\left(x-2\right)^2+4\left(x+3\right)^2\)

\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2-4x+4\right)+4\left(x^2+6x+9\right)\)

\(=x^2+2x^2+4x+2+3x^2-12x+12+4x^2+24x+36\)

\(=10x^2+16x+50\)