Question

Представьте в виде многочлена
(2а-3b)^2
(xy+1/2y)^2
(5m^2-2n^3)^2
(0,2p^3+3q^4)^2




(y+x)^3
(m-n)^3
(5+a)^3
(3a-b)^2
(10-x^2)^3
(x^2+y^2)^3

Answer 1

${(2a - 3b)}^{2} = 4 {a}^{2} - 12ab + 9 {b}^{2} \ {(xy + frac{1}{2} y)}^{2} = {x}^{2} {y}^{2} + x {y}^{2} + frac{1}{4} {y}^{2} \ {(5 {m}^{2} - 2 {n}^{3} )}^{2} = 25 {m}^{4} - 20 {m}^{2} {n}^{3} + 4 {n}^{6}$
${(0.2 {p}^{3} + 3 {q}^{4}) }^{2} = 0.04 {p}^{6} + 1.2 {p}^{3} {q}^{4} + 9 {q}^{8}$

${(y + x)}^{3} = {y}^{3} + 3x {y}^{2} + 3 {x}^{2} y + {x}^{3} \ {(m - n)}^{3} = {m}^{3} - 3 {m}^{2} n + 3m {n}^{2} - {n}^{3}$
${(5 + a)}^{3} = 125 + 75a + 15{a}^{2} + {a}^{3} \ {(3a - b)}^{2} = 9 {a}^{2} - 6ab + {b}^{2} \ {(10 - {x}^{2} )}^{3} = 1000 - 300 {x}^{2} + 30 {x}^{4} - {x}^{6}$
${( {x}^{2} + {y}^{2} )}^{3} = {x}^{6} + 3 {x}^{4} {y}^{2} + 3 {x}^{2} {y}^{4} + {y}^{6}$