Упр.123 Вариант 3 Дидактические материалы ГДЗ Мерзляк Полонский 7 класс (Алгебра)

1) x^2-100; 6) m^8-n^10;
2) 36-81b^2; 7) 0,16p^4-q^6;
3) 9x^2-64y^2; 8) 1,21z^8-225t^14;
4) 0,09a^2-1,44b^2; 9)-4+169x^4 y^18;
5) x^4 y^4-9/16; 10) 2 14/25 x^4 y^4-1 17/64 a^6 b^8.

1. $$x^2-100=x^2-10^2=(x-10)(x+10)$$
2. $$36-81b^2=6^2-(9b)^2=(6-9b)(6+9b)$$
3. $$9x^2-64y^2=(3x)^2-(8y)^2=(3x-8y)(3x+8y)$$
4. $$0{,}09a^2-1{,}44b^2=(0{,}3a)^2-(1{,}2b)^2=(0{,}3a-1{,}2b)(0{,}3a+1{,}2b)$$
5. $$x^4y^4-\frac{9}{16}=(x^2y^2)^2-\left(\frac{3}{4}\right)^2=\left(x^2y^2-\frac{3}{4}\right)\left(x^2y^2+\frac{3}{4}\right)$$
6. $$m^8-n^{10}=(m^4)^2-(n^5)^2=(m^4-n^5)(m^4+n^5)$$
7. $$0{,}16p^4-q^6=(0{,}4p^2)^2-(q^3)^2=(0{,}4p^2-q^3)(0{,}4p^2+q^3)$$
8. $$1{,}21z^8-225t^{14}=(1{,}1z^4)^2-(15t^7)^2=(1{,}1z^4-15t^7)(1{,}1z^4+15t^7)$$
9. $$-4+169x^4y^{18}=169x^4y^{18}-4=(13x^2y^9)^2-2^2=(13x^2y^9-2)(13x^2y^9+2)$$
10. $$2\frac{14}{25}x^4y^4-1\frac{17}{64}a^6b^8=\frac{64}{25}x^4y^4-\frac{81}{64}a^6b^8$$
    $$=\left(\frac{8}{5}x^2y^2\right)^2-\left(\frac{9}{8}a^3b^4\right)^2$$
    $$=\left(\frac{8}{5}x^2y^2-\frac{9}{8}a^3b^4\right)\left(\frac{8}{5}x^2y^2+\frac{9}{8}a^3b^4\right)$$

Ответ

1) $$(x-10)(x+10)$$

2) $$(6-9b)(6+9b)$$

3) $$(3x-8y)(3x+8y)$$

4) $$(0{,}3a-1{,}2b)(0{,}3a+1{,}2b)$$

5) $$\left(x^2y^2-\frac{3}{4}\right)\left(x^2y^2+\frac{3}{4}\right)$$

6) $$(m^4-n^5)(m^4+n^5)$$

7) $$(0{,}4p^2-q^3)(0{,}4p^2+q^3)$$

8) $$(1{,}1z^4-15t^7)(1{,}1z^4+15t^7)$$

9) $$(13x^2y^9-2)(13x^2y^9+2)$$

10) $$\left(\frac{8}{5}x^2y^2-\frac{9}{8}a^3b^4\right)\left(\frac{8}{5}x^2y^2+\frac{9}{8}a^3b^4\right)$$