Question

(4х-7)²=
(6х+10)²=
25-а²=
9а-¼=
36m-49n²=
7a²-7=
6x=1
3x+7=0
4y+15=6y+17

Answer 1

Ответ:

$(4x-7)^2=16x^2-56x+49\\\\(6x+10)^2=36x^2+120x+100\\\\25-a^2=(5-a)(5+a)\\\\9a^{-\frac{1}{4}}=\dfrac{9}{\sqrt[4]{a}}\\\\9a-\dfrac{1}{4}=(3\sqrt{a}-\dfrac{1}{2})(3\sqrt{a}+\dfrac{1}{2})\\\\36m-49n^2=(6\sqrt{m}-7n)(6\sqrt{m}+7n)\\\\7a^2-7=7(a-1)(a+1)\\\\6x=1\ \ ,\ \ x=\dfrac{1}{6}\\\\3x+7=0\ \ ,\ \ 3x=-7\ \ ,\ \ x=-\dfrac{7}{3}\ \ ,\ \ x=-2\dfrac{1}{3}\\\\4y+15=6y+17\ \ ,\ \ 2y=-2\ \ ,\ \ x=1$