Question

Помогите пожалуйста дам лучший ответ​

Answer 1

Ответ:

$1)\ \ 16x^2+\ \boxed{*}\ +y^2=\underbrace {(4x)^2+\ \boxed {*}\ +y^2}_{a^2+2ab+b^2}=(4x)^2+\underbrace {2\cdot 4x\cdot y}_{\boxed {*}=2ab}+y^2=\\\\=16x^2+8xy+y^2=(4x+y)^2\\\\{}\boxed{*}=8xy\\\\\\\\2)\ \ \underbrace{49p^2-14p+\boxed{*}}_{a^2-2ab+b^2}=(7p)^2-2\cdot 7p\cdot 1+\underbrace{1^2}_{\boxed{*}=b^2}=(7p-1)^2\\\\\boxed{*}=1\\\\\\\\3)\ \ \underbrace{25-10a+\boxed{*}}_{x^2-2xy+y^2}=5^2-2\cdot 5\cdot a+\underbrace{a^2}_{\boxed{*}=y^2}=(5-a)^2\\\\\boxed{*}=a^2$

$4)\ \ \underbrace{\boxed{*}-36ab+\boxed{**}}_{x^2-2xy+y^2}=\underbrace{(2a)^2}_{\boxed{*}=x^2}-2\cdot 2a\cdot 9b+\underbrace{(9b)^2}_{\boxed{**}=y^2}=4a^2-36ab+81b^2=\\\\\\=(2a-9b)^2\\\\\boxed{*}=4a^2\ \ ,\ \ \boxed{**}=81b^2\\\\\\\\ili\ \ \underbrace{\boxed{*}-36ab+\boxed{**}}_{x^2-2xy+y^2}=\underbrace{(2b)^2}_{\boxed{*}=x^2}-2\cdot 2b\cdot 9a+\underbrace{(9a)^2}_{\boxed{**}=y^2}=4b^2-36ab+81a^2=\\\\\\=(2b-9a)^2\\\\\boxed{*}=4b^2\ \ ,\ \ \boxed{**}=81a^2$