№32.21(1,4,6) и №32.22(1,2,4) срочно даю 60 баллов!!!!!!!!!!!. Created by kolia0. algebra-ru

Объяснение:32.21.1) 9x²-48xy+64y²4) 25k²+12kt+1,44t²6) 32.22.1) 1,69+10,4m²n²+162) 6,25+20a²b+16b²4)

№32.21(1,4,6) и №32.22(1,2,4) срочно даю 60 баллов!!!!!!!!!!!...

LINDINDRIJ

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Объяснение:

32.21.

1) 9x²-48xy+64y²

4) 25k²+12kt+1,44t²

6) $\frac{49}{64}c^{2}-cd+\frac{16}{49}d^{2}$

32.22.

1) 1,69 $m^{4}$ +10,4m²n²+16 $n^{4}$

2) 6,25 $a^{4}$ +20a²b+16b²

4) $5,76m^{6}-3,6m^{3} t +\frac{9}{16}t^{2}$

3 votes Thanks 1

NNNLLL54

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Ответ:

$\boxed {\ (a\pm b)^2=a^2\pm 2ab+b^2\ }$

$1)\ \ (3x-8y)^2=9x^2-48xy+64y^2\\\\4)\ \ (5k+1,2\, t)^2=25k^2+12\, kt+1,44\, t^2\\\\6)\ \ \Big(\dfrac{7}{8}\, c-\dfrac{4}{7}\, d \Big)^2=\dfrac{49}{64}\, c^2-c\, d+\dfrac{16}{49}\, d^2\\\\\\1)\ \ (1,3m^2+4n^2)^2=1,69m^4+10,4\, m^2\, n^2+16\, n^4\\\\2)\ \ (2,5a^2+4b)^2=6,25a^4+20\, a^2\, b+16b^2\\\\4)\ \ \Big(2,4m^3-\dfrac{3}{4}\, t\Big)^2=5,76\, m^6-3,6\, m^3\, t+\dfrac{9}{16}\, t^2$

6 votes Thanks 6

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