Question

ДАЮ 80БАЛОВ!ПОЖАЛУЙСТА ПОМОГИТЕ СРОЧНО!
подайте у вигляді добутку 1)y²-9 2)25a² 3)1/16-m⁴

Answer 1

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Answer 2

Ответ:

Применяем формулы разности квадратов :  $\bf a^2-b^2=(a-b)(a+b)$  ,

а также суммы и разности кубов :   $\bf a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$  .

$\bf 467)\ \ 1)y^2-9=y^2-3^2=(y-3)(y+3)\\\\2)\ \ 25a^2-49=(5a)^2-7^2=(5a-7)(5a+7)\\\\3)\ \ \dfrac{1}{16}-m^4=\Big(\dfrac{1}{4}\Big)^2-\Big(m^2\Big)^2=\Big(\dfrac{1}{4}-m^2\Big)\Big(\dfrac{1}{4}+m^2\Big)=\\\\=\Big(\dfrac{1}{2}-m\Big)\Big(\dfrac{1}{2}+m\Big)\Big(\dfrac{1}{4}+m^2\Big)\\\\4)\ \ b^6-0,04=(b^3)^2-(0,2)^2=(b^3-0,2)(b^3+0,2)=\\\\=(b-\sqrt[3]{\bf 0,2})(b^2+b\sqrt[3]{\bf 0,2}+\sqrt[3]{\bf 0,04})(b+\sqrt[3]{\bf 0,2} )(b-b\sqrt[3]{\bf 0,2}+\sqrt[3]{\bf 0,04})$

$\bf 5)\ \ 64x^2-81m^4y^2=(8x-8m^2y)(8x+9m^2y)$  

$\bf 6)\ \ x^{10}-y^{12}=(x^5-y^6)(x^5+y^6)$    

$\boldsymbol{469)\ \ 1)\ \ 36a^2c^2-25=(6ac-5)(6ac+5)}\\\\\boldsymbol{2)\ \ 49-m^2n^2=(7-mn)(7+mn})\\\\\boldsymbol{4)\ \ 100u^2v^2-121}\cdot l^2\boldsymbol{=(10uv-11}\cdot l\, \boldsymbol{)(10uv+11}\cdot l\, \boldsymbol{)}\\\\\bf 5)\ \ m^{12}-c^4=(m^6-c^2)(m^6+c^2)=(m^3-c)(m^3+c)(m^6+c^2)$  

$\bf 471)\ \ 1)\ \ p^8-16t^6=(p^4)^2-(4t^3)^2=(p^4-4t^3)(p^4+4t^3)\\\\2)\ \ 36\, t^8-49u^{10}=(6t^4)^2-(7u^5)^2=(6t^4-7u^5)(6t^4+7u^5)\\\\3)\ \ 0,25-p^{12}c^2=(0,5)^2-(p^6c)^2=(0,5-p^6c)(0,5+p^6c)\\\\4)\ \ 4m^2-121u^8c^{10}=(2m)^2-(11u^4c^5)^2=(2m-11u^4c^5)(2m+11u^4c^5)\\\\5)\ \ -\dfrac{25}{36}m^2n^4+\dfrac{36}{49}p^4q^6=\Big(\dfrac{6}{7}p^2q^3-\dfrac{5}{6}mn^2\Big)\Big(\dfrac{6}{7}p^2q^3+\dfrac{5}{6}mn^2\Big)$

$\bf 6)\ \ 2\dfrac{1}{4}m^4n^6-1\dfrac{9}{16}p^8q^{10}=\dfrac{9}{4}m^4n^6-\dfrac{25}{16}p^8q^{10}=\\\\=\Big(\dfrac{3}{2}m^2n^3-\dfrac{5}{4}p^4q^5\Big)\Big(\dfrac{3}{2}m^2n^3+\dfrac{5}{4}p^4q^5\Big)$  

$\bf 474)\ \ 42^2-32^2=(42-32)(42+32)=10\cdot 74=740\\\\25^2-75^2=(25-75)(25+75)=-50\cdot 100=-5000\\\\\dfrac{260}{25^2-1}=\dfrac{260}{(25-1)(25+1)}=\dfrac{260}{24\cdot 26}=\dfrac{260}{624}=\dfrac{65}{156}\\\\\\\dfrac{26^2-12^2}{50^2-12^2}=\dfrac{(26-12)(26+12)}{(50-12)(50+12)}=\dfrac{14\cdot 38}{38\cdot 62}=\dfrac{7}{31}$