Question

представьте в виде многочлена стандартного вида:
а)(5k-3y)²;б)(5k+4/5 k^5)²;в)(5а-12c)(12c+5a)

Answer 1

$(5k-3y)^2=(5k)^2-2*5k*3y+(3y)^2=5^2k^2-30ky+3^2y^2=$
$=25k^2-30ky+9y^2$

$(5k+ frac{4}{5}k^5 )^2=(5k)^2+2*5k* frac{4}{5}k^5+( frac{4}{5}k^5 )^2=$
$=5^2k^2+8k*k^5+frac{4^2}{5^2}k^{5*2}= 25k^2+8k^6+frac{16}{25}k^{10}$

$(5a-12c)(12c+5a)=(5a-12c)(5a+12c)=(5a)^2-(12c)^2=$
$=5^2a^2-12^2c^2=25a^2-144c^2$