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

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^2 y^3-1/5 ay^2 )(7a^2 y^3+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+2/9 b^3 )(2/9 b^3-1,3a^11 ).

1. $$ (a+7)(a-7)=a^2-7^2=a^2-49 $$
2. $$ (6+x)(x-6)=(x+6)(x-6)=x^2-6^2=x^2-36 $$
3. $$ (4b-1)(4b+1)=(4b)^2-1^2=16b^2-1 $$
4. $$ (8m+3y)(3y-8m)=(3y+8m)(3y-8m)=(3y)^2-(8m)^2=9y^2-64m^2 $$
5. $$ (x^7-q^5)(x^7+q^5)=(x^7)^2-(q^5)^2=x^{14}-q^{10} $$
6. $$ \left(7a^2y^3-\frac15 ay^2\right)\left(7a^2y^3+\frac15 ay^2\right)=\left(7a^2y^3\right)^2-\left(\frac15 ay^2\right)^2 $$
   $$ =49a^4y^6-\frac1{25}a^2y^4 $$
7. $$ (0{,}3p^3+0{,}2q^4)(0{,}3p^3-0{,}2q^4)=(0{,}3p^3)^2-(0{,}2q^4)^2 $$
   $$ =0{,}09p^6-0{,}04q^8 $$
8. $$ (x^4-y^4)(x^4+y^4)(x^8+y^8)=\bigl((x^4)^2-(y^4)^2\bigr)(x^8+y^8) $$
   $$ =(x^8-y^8)(x^8+y^8)=(x^8)^2-(y^8)^2=x^{16}-y^{16} $$
9. $$ (m^6-n^5)(-m^6-n^5)=-(m^6+n^5)(m^6-n^5) $$
   $$ =-\bigl((m^6)^2-(n^5)^2\bigr)=-(m^{12}-n^{10})=n^{10}-m^{12} $$
10. $$ \left(1{,}3a^{11}+\frac29 b^3\right)\left(\frac29 b^3-1{,}3a^{11}\right)=\left(\frac29 b^3+1{,}3a^{11}\right)\left(\frac29 b^3-1{,}3a^{11}\right) $$
    $$ =\left(\frac29 b^3\right)^2-\left(1{,}3a^{11}\right)^2=\frac4{81}b^6-1{,}69a^{22} $$

Ответ

1) $$a^2-49$$; 2) $$x^2-36$$; 3) $$16b^2-1$$; 4) $$9y^2-64m^2$$; 5) $$x^{14}-q^{10}$$; 6) $$49a^4y^6-\frac1{25}a^2y^4$$; 7) $$0{,}09p^6-0{,}04q^8$$; 8) $$x^{16}-y^{16}$$; 9) $$n^{10}-m^{12}$$; 10) $$\frac4{81}b^6-1{,}69a^{22}$$.