Question

ХЕЛП СРОЧНО Представь квадрат двучлена в виде многочлена

Answer 1

Ответ:

$\dfrac{1}{16}z^{4}-\dfrac{9}{20}z^{2}+\dfrac{81}{100}$

Объяснение:

$\bigg (\dfrac{1}{4}z^{2}-\dfrac{9}{10} \bigg )^{2}=\bigg (\dfrac{1}{4}z^{2} \bigg )^{2}-2 \cdot \dfrac{1}{4}z^{2} \cdot \dfrac{9}{10}+ \bigg (\dfrac{9}{10} \bigg )^{2}=\bigg (\dfrac{1}{4} \bigg )^{2} \cdot (z^{2})^{2}-\dfrac{9}{20}z^{2}+\dfrac{9^{2}}{10^{2}}=$

$=\dfrac{1^{2}}{4^{2}}z^{2 \cdot 2}-\dfrac{9}{20}z^{2}+\dfrac{81}{100}=\dfrac{1}{16}z^{4}-\dfrac{9}{20}z^{2}+\dfrac{81}{100};$

Answer 2

$\displaystyle (\frac{1}{4}z^2+\frac{9}{10})^2=(\frac{1}{4}z^2)^2+2*\frac{1}{4}z^2*\frac{9}{10}+(\frac{9}{10})^2=\frac{1}{16}z^4-\frac{9}{20}z^2+\frac{81}{100}$