Question

4 столбик срочно даю 30 баллов

Answer 1

Ответ:

Смотри решение на фото

Answer 2

$\displaystyle \tt x^{10}-y^8=x^{2\cdot5}-y^{2\cdot4}=(x^5)^2-(y^4)^2=\boxed{\bold{(x^5-y^4)(x^5+y^4)}}$

$\displaystyle \tt 0,04x^4-0,25y^2=\frac{1}{25}x^4-\frac{1}{4}y^2=\frac{1}{100}(4x^4-25y^2)=\frac{1}{100}(2x^2-5y)(2x^2+5y)=\boxed{\bold{0,01(2x^2-5y)(2x^2+5y)}}$

$\displaystyle \tt \frac{4}{9}m^2-\frac{9}{16}n^4=\frac{1}{144}\cdot(64m^2-81n^2)=\boxed{\bold{\frac{1}{144}(8m-9n)(8m+9n)}}$

$\displaystyle \tt 1,69y^{14}-1,21=\frac{169}{100}y^{14}-\frac{121}{100}=\frac{1}{100}(169y^{14}-121)=\frac{1}{100}(13y^7-11)(13y^7+11)=\boxed{\bold{0,01(13y^7-11)(13y^7+11)}}$

$\displaystyle \tt 121m^8n^8-9=11^2m^{2\cdot4}n^{2\cdot4}-3^2=11^2\cdot(m^4)^2\cdot(n^4)^2-3^2=(11m^4n^4)^2-3^2=\boxed{\bold{(11m^4n^4-3)(11m^4n^4+3)}}$