Question

1. Разложите на множители: а) 2,25a²-0,4962; в) -1,44x y8+25; д) 9х2+у2+6ху-6cx-2cy; ж) 2 ; a12+b9; 10 27 7 6) 1 m2-1 9 г) е) - 3) 13 36 36(u+1)2-25(2u-1)2; 15 nt; +1; m2+5m+5n-n². 8 ПОМОГИТЕ ПЖ СРОЧНО НАДО ​

Answer 1

Ответ:

Объяснение:

1. a)

$2,25a^2-0,49b^2=(1.5a)^2-(0.7b)^2=(1.5a-0.7b)(1.5a+0.7b)$

б)

$1\frac{7}{9} m^2-1\frac{13}{36} n^4=\frac{16}{9} m^2-\frac{49}{36} n^4=(\frac{4}{3} m)^2-(\frac{7}{6} n^2)^2=(\frac{4}{3}m-\frac{7}{6}n^2)(\frac{4}{3}m+\frac{7}{6}n^2)$

в)

$-1.44x^6y^8+25=-(1.2x^3y^4)^2+5^2=5^2-(1.2x^3y^4)^2=(5-1.2x^3y^4)(5+1,2x^3y^4)$

г)

$36(u+1)^2-25(2u-1)^2=(6(u+1))^2-(5(2u-1))^2=(6u+6)^2-(10u-5)^2=(6u+6-10u+5)(6u+6+10u-5)=(11-4u)(16u+1)$

д)

$9x^2+y^2+6xy-6cx-2cy=(3x)^2+y^2+2*3x*y-2c(3x+y)=(3x+y)^2-2c(3x+y)=(3x+y)(3x+y-2c)$

е)

$-\frac{p^{15}}{8} +1=1^3-\frac{(p^5)^3}{2^3} =1^3-(\frac{p^5}{2} )^3=(1-\frac{p^5}{2} )(1 + \frac{p^5}{2} + \frac{p^{10}}{4} )$

ж)

$2\frac{10}{27} a^{12}+b^9=\frac{64}{27} a^{12}+b^9=(\frac{4}{3} a^4)^3+(b^3)^3=(\frac{4}{3} a^4+b^3)((\frac{4}{3} a^4)^2-\frac{4}{3} a^4b^3+(b^3)^2)=(\frac{4}{3} a^4+b^3)(\frac{16}{9} a^8-\frac{4}{3} a^4b^3+b^6)$

з)

$m^2+5m+5n-n^2=m^2-n^2+5m+5n=(m-n)(m+n)+5(m+n)=(m+n)(m-n+5)$