Контрольные задания >
2) \(\frac{9}{25} + \frac{6}{5}b + b^{2};\)
4) \(\frac{100}{121}k^{2} - \frac{20}{11}tk + t^{2};\)
6) \(\frac{400}{441}t^{2} + \frac{40}{21}nt + n^{2}.\)
2) \(\frac{9}{16} - \frac{3}{2}y + y^{2};\)
4) \(n^{2} - \frac{9}{4}cn + \frac{81}{64}c^{2};\)
6) \(t^{2} - \frac{17}{5}dt + \frac{289}{100}d^{2}.\)
Вопрос:
2) \(\frac{9}{25} + \frac{6}{5}b + b^{2};\)
4) \(\frac{100}{121}k^{2} - \frac{20}{11}tk + t^{2};\)
6) \(\frac{400}{441}t^{2} + \frac{40}{21}nt + n^{2}.\)
2) \(\frac{9}{16} - \frac{3}{2}y + y^{2};\)
4) \(n^{2} - \frac{9}{4}cn + \frac{81}{64}c^{2};\)
6) \(t^{2} - \frac{17}{5}dt + \frac{289}{100}d^{2}.\)
Смотреть решения всех заданий с листаОтвет:
-
$$\frac{9}{25} + \frac{6}{5}b + b^{2} = (\frac{3}{5})^{2} + 2 \cdot \frac{3}{5} \cdot b + b^{2} = (\frac{3}{5} + b)^{2}$$
Ответ: $$(\frac{3}{5} + b)^{2}$$
-
$$\frac{100}{121}k^{2} - \frac{20}{11}tk + t^{2} = (\frac{10}{11}k)^{2} - 2 \cdot \frac{10}{11}k \cdot t + t^{2} = (\frac{10}{11}k - t)^{2}$$
Ответ: $$(\frac{10}{11}k - t)^{2}$$
-
$$\frac{400}{441}t^{2} + \frac{40}{21}nt + n^{2} = (\frac{20}{21}t)^{2} + 2 \cdot \frac{20}{21}t \cdot n + n^{2} = (\frac{20}{21}t + n)^{2}$$
Ответ: $$(\frac{20}{21}t + n)^{2}$$
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$$\frac{9}{16} - \frac{3}{2}y + y^{2} = (\frac{3}{4})^{2} - 2 \cdot \frac{3}{4} \cdot y + y^{2} = (\frac{3}{4} - y)^{2}$$
Ответ: $$(\frac{3}{4} - y)^{2}$$
-
$$n^{2} - \frac{9}{4}cn + \frac{81}{64}c^{2} = n^{2} - 2 \cdot n \cdot \frac{9}{8}c + (\frac{9}{8}c)^{2} = (n - \frac{9}{8}c)^{2}$$
Ответ: $$(n - \frac{9}{8}c)^{2}$$
-
$$t^{2} - \frac{17}{5}dt + \frac{289}{100}d^{2} = t^{2} - 2 \cdot t \cdot \frac{17}{10}d + (\frac{17}{10}d)^{2} = (t - \frac{17}{10}d)^{2}$$
Ответ: $$(t - \frac{17}{10}d)^{2}$$
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