Упр.120 Вариант 2 Дидактические материалы ГДЗ Мерзляк Полонский 7 класс (Алгебра)
Представьте в виде многочлена выражение:
- $$(x-6)(x+6)$$;
- $$(3+x)(x-3)$$;
- $$(3b-5)(3b+5)$$;
- $$(5x+8y)(8y-5x)$$;
- $$(m^5-n^3)(m^5+n^3)$$;
- $$\left(5a^2b-\frac{1}{4}ab^2\right)\left(5a^2b+\frac{1}{4}ab^2\right)$$;
- $$(0{,}5x^3+0{,}2y^4)(0{,}5x^3-0{,}2y^4)$$;
- $$(a^5-b^5)(a^5+b^5)(a^{10}+b^{10})$$;
- $$(-x^7-y^3)(y^3-x^7)$$;
- $$\left(\frac{2}{3}y^6+1{,}2x^{11}\right)\left(1{,}2x^{11}-\frac{2}{3}y^6\right)$$.
1) $$\left(x-6\right)\left(x+6\right)=x^2-6^2=x^2-36;$$
2) $$\left(3+x\right)\left(x-3\right)=\left(x+3\right)\left(x-3\right)=x^2-3^2=x^2-9;$$
3) $$\left(3b-5\right)\left(3b+5\right)=\left(3b\right)^2-5^2=9b^2-25;$$
4) $$\left(5x+8y\right)\left(8y-5x\right)=\left(8y+5x\right)\left(8y-5x\right)=\left(8y\right)^2-\left(5x\right)^2=64y^2-25x^2;$$
5) $$\left(m^5-n^3\right)\left(m^5+n^3\right)=\left(m^5\right)^2-\left(n^3\right)^2=m^{10}-n^6;$$
6) $$\left(5a^2b-\frac14ab^2\right)\left(5a^2b+\frac14ab^2\right)=\left(5a^2b\right)^2-\left(\frac14ab^2\right)^2=25a^4b^2-\frac1{16}a^2b^4;$$
7) $$\left(0{,}5x^3+0{,}2y^4\right)\left(0{,}5x^3-0{,}2y^4\right)=\left(0{,}5x^3\right)^2-\left(0{,}2y^4\right)^2=0{,}25x^6-0{,}04y^8;$$
8) $$\left(a^5-b^5\right)\left(a^5+b^5\right)\left(a^{10}+b^{10}\right)=\left(a^{10}-b^{10}\right)\left(a^{10}+b^{10}\right)=a^{20}-b^{20};$$
9) $$\left(-x^7-y^3\right)\left(y^3-x^7\right)=-\left(y^3+x^7\right)\left(y^3-x^7\right)=-\left(\left(y^3\right)^2-\left(x^7\right)^2\right)=x^{14}-y^6;$$
10) $$\left(\frac23y^6+1{,}2x^{11}\right)\left(1{,}2x^{11}-\frac23y^6\right)=\left(1{,}2x^{11}+\frac23y^6\right)\left(1{,}2x^{11}-\frac23y^6\right)=\left(1{,}2x^{11}\right)^2-\left(\frac23y^6\right)^2=1{,}44x^{22}-\frac49y^{12}.$$
