Упр.463 ГДЗ Никольский Потапов 7 класс (Алгебра)

б) 25×2-1;
в) 1/4*m2-16n2;
г) 100a2-0,25b2;
д) x12-y2;
е) m6-n6;
ж) 2*1/4-c4;
з) 1*9/16 *a10 — 0,01b2;
и) x4-y4.

1. $$9a^2-4=(3a)^2-2^2=(3a-2)(3a+2).$$
2. $$25x^2-1=(5x)^2-1^2=(5x-1)(5x+1).$$
3. $$\frac14 m^2-16n^2=\left(\frac12 m\right)^2-(4n)^2=\left(\frac12 m-4n\right)\left(\frac12 m+4n\right).$$
4. $$100a^2-0{,}25b^2=(10a)^2-(0{,}5b)^2=(10a-0{,}5b)(10a+0{,}5b).$$
5. $$x^{12}-y^2=(x^6)^2-y^2=(x^6-y)(x^6+y).$$
6. $$m^6-n^6=(m^3)^2-(n^3)^2=(m^3-n^3)(m^3+n^3),$$
   $$m^3-n^3=(m-n)(m^2+mn+n^2),$$
   $$m^3+n^3=(m+n)(m^2-mn+n^2).$$
   Тогда
   $$m^6-n^6=(m-n)(m^2+mn+n^2)(m+n)(m^2-mn+n^2).$$
7. $$2\frac14-c^4=\frac94-c^4=\left(\frac32\right)^2-(c^2)^2=\left(\frac32-c^2\right)\left(\frac32+c^2\right).$$
8. $$1\frac{9}{16}a^{10}-0{,}01b^2=\frac{25}{16}a^{10}-0{,}01b^2=\left(\frac54a^5\right)^2-(0{,}1b)^2,$$
   $$\left(\frac54a^5\right)^2-(0{,}1b)^2=\left(\frac54a^5-0{,}1b\right)\left(\frac54a^5+0{,}1b\right).$$
9. $$x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2).$$

Ответ

а) $$(3a-2)(3a+2)$$
б) $$(5x-1)(5x+1)$$
в) $$\left(\frac12 m-4n\right)\left(\frac12 m+4n\right)$$
г) $$(10a-0{,}5b)(10a+0{,}5b)$$
д) $$(x^6-y)(x^6+y)$$
е) $$(m-n)(m+n)(m^2+mn+n^2)(m^2-mn+n^2)$$
ж) $$\left(\frac32-c^2\right)\left(\frac32+c^2\right)$$
з) $$\left(\frac54a^5-0{,}1b\right)\left(\frac54a^5+0{,}1b\right)$$
и) $$(x-y)(x+y)(x^2+y^2)$$