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

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)  \(x^2+6x+9=\left(x+3\right)^2\)

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

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

a, \(\left(x+3\right)^2\)

b,\(\left(x+\frac{1}{2}\right)^2\)Mik giải thích tí nè, cái này =\(x^2+2.x.\frac{1}{2}+\frac{1}{4}\)=\(x^2+x+\frac{1}{4}\)

c,thì mik chịu.

a) x²+6x+9

=x²+2.x.3+3²

=(x+3)²

b) x²+x+\(\frac{1}{4}\)

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

=(x+\(\frac{1}{2}\))²

c) 2xy²+x²y⁴+1

=(xy²)²+2.xy².1+1²

=(xy²+1)²

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

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 + 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

\(2xy^2+x^2y^4+1\)

\(=\left(xy^2\right)^2+2.xy^2+1^2\)

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

làm sai bét