1: y^2-8y+16=(y-4)^2
2: x^2+2x+1=(x+1)^2
3: 9x^2+12x+4=(3x+2)^2
4: 9-12y+4y^2=(3-2y)^2
5: 1/4x^2+7x+49=(1/2x+7)^2
6: x^2-10x+25=(x-5)^2
\(1,y^2-8y+16\\ =y^2-2.4.y+4^2\\ =\left(y-4\right)^2\\ 2,1+2x+x^2=\left(1+x\right)^2\\ 3,9x^2+12x+4=\left(3x\right)^2+2.3x.2+2^2\\ =\left(3x+2\right)^2\\ 4,9-12y+4y^2\\ =\left(3-2y\right)^2\\ 5,49+7x+\dfrac{1}{4}x^2\\ =7^2+2.\dfrac{1}{2}x.7+\left(\dfrac{1}{2}x\right)^2\\ =\left(7+\dfrac{1}{2}x\right)^2\\ 6,x^2-10x+25=\left(x-5\right)^2\)
\(1.\\ y^2-8y+16\\ =y^2-2.y.4+4^2\\ =\left(y-4\right)^2\)
\(2.\\ 1+2x+x^2\\ =1^2+2.1.x+x^2\\ =\left(1+x\right)^2\)
\(3.\\ 9x^2+12x+4\\ =\left(3x\right)^2+2.3x.2+2\\ =\left(3x+2\right)^2\)
\(4.\\ 9-12y+4y^2\\ =3^2-2.3.2y+\left(2y\right)^2\\ =\left(3-2y\right)^2\)
\(5.\\ 49+7x+\dfrac{1}{4}x^2\\ =7^2+2.7.\dfrac{1}{2}+\left(\dfrac{1}{2}x\right)^2\\ =\left(7+\dfrac{1}{2}\right)^2\)
\(6.\\ x^2-10x+25\\ =x^2-2.x.5+5^2\\=\left(x-5\right)^2 \)
