Question

2.Представьте в виде многочлена:

Answer 1

Объяснение:

(х+3)²=х²+6х+9

(4–у)²=16–8у+у²

${(a + \frac{1}{2} b)}^{2} = {a}^{2} + ab + \frac{ 1 }{4} {b}^{2}$

(2m–5)²=4m²–20m+25

(7a+6b)²=49a²+84ab+36b²

(0,2x–10y)²=0,04x²–4xy+100y²

${(9m + \frac{1}{3} n)}^{2} = 81 {m}^{2} + 6mn + \frac{1}{9} {n}^{2}$

(a²–1)²=a⁴–2a+1

(x³–x²)²=(x^6)–2(x^5)+x⁴

(p²+p⁴)²=p⁴+2(p^6)+(p^8)

(–11b+2(a^5))²= 4(a^10)–44(a^5)b+121b²

(–8–4c)²=64+64c+16c²

${(1 \frac{2}{3}p + 2 \frac{2}{5} q) }^{2} = {( \frac{5}{3}p + \frac{12}{5} q) }^{2} = \frac{25}{9} {p}^{2} + \frac{120}{15} pq + \frac{144}{25} {q}^{2} = 2 \frac{7}{9} {p}^{2} +8pq + 5 \frac{19}{25} {q}^{2}$

(12xy²–x²y)²=144x²y⁴–24x³y³+x⁴y²

(4(a^6)+3a⁴b³)²=16(a^12)+24(a^10)b³+9(a^8)(b^6)