A)\(1-2x+x^2\)

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

B)\(4y+4+y^2\)

\(=2^2+4y+y^2\)

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

C)\(\frac{1}{16}+\frac{1}{2}x+x^2\)

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

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

D)\(36x^2+12xy+y^2\)

\(=\left(6x+y\right)^2\)

a) Ta có: \(\left(x^2+9x+18\right)^2+2\left(x^2+9x\right)+37\)

\(=\left(x^2+9x+18\right)^2+2\cdot\left(x^2+9x+18\right)-36+37\)

\(=\left(x^2+9x+19\right)^2\)

b) Ta có: \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

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

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

c) Ta có: \(x^2-2x\left(y+2\right)+y^2+4y+4\)

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

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

d) Ta có: \(x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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

này mình có vài câu không làm được, xin lỗi bạn nha

\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)

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

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

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: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

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

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

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

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

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

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

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

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

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