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

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

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

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

\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)

Bài 1 em dùng HĐT nha

Bài 2:

a. x² + 4x + 4

= x² + 2.2.x + 2²

= (x + 2)²

b. 9x² - 12x + 4

= (3x)² - 3x.2.2 + 2²

= (3x - 2)²

c. \(\dfrac{x^2}{4}+x+1\)

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

= \(\left(\dfrac{x}{2}+1\right)^2\) 

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

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

\(25+10x+x^2=5^2+2.5x+x^2=\left(5+x\right)^2\)

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

a,(x-3)^2

b,(1/4x+2b^2)^2

c,(5+x)^2

d,(1/3-y^4)^2

sorry ,phần b fải là (1/2a+2b^2)^2

a:Sửa đề: \(\dfrac{1}{4}a^2+2ab+4b^2\)

\(=\left(\dfrac{1}{2}a\right)^2+2\cdot\dfrac{1}{2}a\cdot2b+\left(2b\right)^2\)

\(=\left(\dfrac{1}{2}a+2b\right)^2\)

b: Sửa đề:\(y^4-\dfrac{1}{3}y^4+\dfrac{1}{36}\)

\(=y^8-2\cdot y^4\cdot\dfrac{1}{6}+\dfrac{1}{36}\)

\(=\left(y^4-\dfrac{1}{6}\right)^2\)

a) \(6x^2y+9+x^4y^2=\left(x^2y+3\right)^2\)

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

c) \(25y^4-10y^2+1=\left(5y^2-1\right)^2\)

\(a,=\left(x^2y+3\right)^2\\ b,=\left(2x+y\right)^2\\ c,=\left(5y^2-1\right)^2\)

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. (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

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