Упр.120 Вариант 3 Дидактические материалы ГДЗ Мерзляк Полонский 7 класс (Алгебра)

Представьте в виде многочлена выражение:

1. 1) $$(a+7)(a-7)$$;
   2) $$(6+x)(x-6)$$;
   3) $$(4b-1)(4b+1)$$;
   4) $$(8m+3y)(3y-8m)$$;
   5) $$(x^7-q^5)(x^7+q^5)$$;
   6) $$(7a^2y^3-\frac{1}{5}ay^2)(7a^2y^3+\frac{1}{5}ay^2)$$;
   7) $$(0{,}3p^3+0{,}2q^4)(0{,}3p^3-0{,}2q^4)$$;
   8) $$(x^4-y^4)(x^4+y^4)(x^8+y^8)$$;
   9) $$(m^6-n^5)(-m^6-n^5)$$;
   10) $$(1{,}3a^{11}+\frac{2}{9}b^3)(\frac{2}{9}b^3-1{,}3a^{11})$$.

1. $$\begin{aligned} (a+7)(a-7)&=a^2-7^2\\ &=a^2-49 \end{aligned}$$
2. $$\begin{aligned} (6+x)(x-6)&=(x+6)(x-6)\\ &=x^2-6^2\\ &=x^2-36 \end{aligned}$$
3. $$\begin{aligned} (4b-1)(4b+1)&=(4b)^2-1^2\\ &=16b^2-1 \end{aligned}$$
4. $$\begin{aligned} (8m+3y)(3y-8m)&=(3y+8m)(3y-8m)\\ &=(3y)^2-(8m)^2\\ &=9y^2-64m^2 \end{aligned}$$
5. $$\begin{aligned} (x^7-q^5)(x^7+q^5)&=(x^7)^2-(q^5)^2\\ &=x^{14}-q^{10} \end{aligned}$$
6. $$\begin{aligned} \left(7a^2y^3-\frac15 ay^2\right)\left(7a^2y^3+\frac15 ay^2\right) &=\left(7a^2y^3\right)^2-\left(\frac15 ay^2\right)^2\\ &=49a^4y^6-\frac1{25}a^2y^4 \end{aligned}$$
7. $$\begin{aligned} (0{,}3p^3+0{,}2q^4)(0{,}3p^3-0{,}2q^4) &=(0{,}3p^3)^2-(0{,}2q^4)^2\\ &=0{,}09p^6-0{,}04q^8 \end{aligned}$$
8. $$\begin{aligned} (x^4-y^4)(x^4+y^4)(x^8+y^8) &=\bigl((x^4)^2-(y^4)^2\bigr)(x^8+y^8)\\ &=(x^8-y^8)(x^8+y^8)\\ &=(x^8)^2-(y^8)^2\\ &=x^{16}-y^{16} \end{aligned}$$
9. $$\begin{aligned} (m^6-n^5)(-m^6-n^5) &=-\,(m^6+n^5)(m^6-n^5)\\ &=-\bigl((m^6)^2-(n^5)^2\bigr)\\ &=-(m^{12}-n^{10})\\ &=n^{10}-m^{12} \end{aligned}$$
10. $$\begin{aligned} \left(1{,}3a^{11}+\frac29 b^3\right)\left(\frac29 b^3-1{,}3a^{11}\right) &=\left(\frac29 b^3+1{,}3a^{11}\right)\left(\frac29 b^3-1{,}3a^{11}\right)\\ &=\left(\frac29 b^3\right)^2-\left(1{,}3a^{11}\right)^2\\ &=\frac{4}{81}b^6-1{,}69a^{22} \end{aligned}$$