Упр.9 ГДЗ Колягин Ткачёва 10 класс (Алгебра)

1) x8 — 4;
2) 25n2 — 49р4;
3) 1*9/16а2 — 0,09b2;
4) 0,0081×6 — 1*7/9y10.

1. $$x^8-4=(x^4)^2-2^2=(x^4-2)(x^4+2).$$
2. $$25n^2-49p^4=(5n)^2-(7p^2)^2=(5n-7p^2)(5n+7p^2).$$
3. $$1\frac{9}{16}a^2-0{,}09b^2=\left(\frac{5}{4}a\right)^2-(0{,}3b)^2=\left(\frac{5}{4}a-0{,}3b\right)\left(\frac{5}{4}a+0{,}3b\right).$$
4. $$0{,}0081x^6-1\frac{7}{9}y^{10}=(0{,}09x^3)^2-\left(\frac{4}{3}y^5\right)^2=\left(0{,}09x^3-\frac{4}{3}y^5\right)\left(0{,}09x^3+\frac{4}{3}y^5\right).$$

Ответ

$$x^8-4=(x^4-2)(x^4+2);$$
$$25n^2-49p^4=(5n-7p^2)(5n+7p^2);$$
$$1\frac{9}{16}a^2-0{,}09b^2=\left(\frac{5}{4}a-0{,}3b\right)\left(\frac{5}{4}a+0{,}3b\right);$$
$$0{,}0081x^6-1\frac{7}{9}y^{10}=\left(0{,}09x^3-\frac{4}{3}y^5\right)\left(0{,}09x^3+\frac{4}{3}y^5\right).$$