Question

Помогите,а то я забыл как решать!!!!

Answer 1

На зеленой доске:
1) a^2-324
2) 121-x^2
3) x^2-36
4) (12-y^2)(12+y^2)
5) (x+1)(x-1)
6) (7x-8y)(7x+8y)
7) 4x^2-12x+9
8) 25y^2+10y+1
9) (x-12)^2
10) (x+1/2)^2
11) (4x+1)^2

На белой доске:
49-x^2
81-x^2
x^2-9
64-y^2
x^2-18x+81
x^2+12x+36
16-8x+x^2
x^4+2x^2+1

Answer 2

$1.$
$(7-x)(7+x)=49-x^{2}$
$(9-x)(x+9)=81-x^{2}$
$(x+3)(x-3)=x^{2}-9$
$(y+8)(8-y)=64-y^{2}$
$(x-9)^{2}=x^{2}-18x+81$
$(x+6)^{2}=x^{2}+12x+36$
$(4-x)^{2}=16-8x+x^{2}$
$(x^{2}+1)^{2}=x^{4}+2x^{2}+1$
$2.$
$(a-18)(a+18)=a^{2}-324$
$(11-x)(x+11)=121-x^{2}$
$(6+x)(x-6)=x^{2}-36$
$144-y^{4}=(12-y^{2})(12+y^{2})$
$x^{2}-1=(x-1)(x+1)$
$49x^{2}-64y^{2}=(7x-8y)(7x+8y)$
$(2x-3)^{2}=4x^{2}-12x+9$
$(5y+1)^{2}=25y^{2}+10y+1$
$3.$
$x^{2}-24x+144=(x-12)^{2}$
$x^{2}+x+frac{1}{4}=(x+frac{1}{2})^{2}$
$16x^{2}+8x+1=(4x+1)^{2}$