Упр.995 ГДЗ Макарычев 7 класс (Алгебра)

а) 1-a2b2;
б) 4x2y4-9;
в) 0,09×6 — 0,49y2;
г) 1,21a2-0,36b6;
д) 1*7×2/9 — 9y2/16;
е) 0,01a2b4-1.

1. $$1-a^2b^2=1-(ab)^2=(1-ab)(1+ab)$$
2. $$4x^2y^4-9=(2xy^2)^2-3^2=(2xy^2-3)(2xy^2+3)$$
3. $$0{,}09x^6-0{,}49y^2=(0{,}3x^3)^2-(0{,}7y)^2=(0{,}3x^3-0{,}7y)(0{,}3x^3+0{,}7y)$$
4. $$1{,}21a^2-0{,}36b^6=(1{,}1a)^2-(0{,}6b^3)^2=(1{,}1a-0{,}6b^3)(1{,}1a+0{,}6b^3)$$
5. $$\frac{7}{9}x^2-\frac{9}{16}y^2=\left(\frac{2\sqrt7}{6}x\right)^2-\left(\frac34y\right)^2$$
   $$=\left(\frac{\sqrt7}{3}x-\frac34y\right)\left(\frac{\sqrt7}{3}x+\frac34y\right)$$
6. $$0{,}01a^2b^4-1=(0{,}1ab^2)^2-1^2=(0{,}1ab^2-1)(0{,}1ab^2+1)$$

Ответ

а) $$\left(1-ab\right)\left(1+ab\right)$$;
б) $$\left(2xy^2-3\right)\left(2xy^2+3\right)$$;
в) $$\left(0{,}3x^3-0{,}7y\right)\left(0{,}3x^3+0{,}7y\right)$$;
г) $$\left(1{,}1a-0{,}6b^3\right)\left(1{,}1a+0{,}6b^3\right)$$;
д) $$\left(\frac{\sqrt7}{3}x-\frac34y\right)\left(\frac{\sqrt7}{3}x+\frac34y\right)$$;
е) $$\left(0{,}1ab^2-1\right)\left(0{,}1ab^2+1\right)$$.