b/ B=(cosa−sina)2+(cosa+sina)2+cos4a−sin4a−2cos2aB=(cosa−sina)2+(cosa+sina)2+cos4a−sin4a−2cos2a

=cos2a−2cosa.sina+sin2a+cos2a+2cosa.sina+sin2a+(cos2a+sin2a)(cos2a−sin2a)−2cos2a=cos2a−2cosa.sina+sin2a+cos2a+2cosa.sina+sin2a+(cos2a+sin2a)(cos2a−sin2a)−2cos2a

=2+(cos2a−sin2a)−2cos2a=2+(cos2a−sin2a)−2cos2a

=2−sin2a−cos2a=2−1=1

Ủa đâu phải đâu bạn

ĐKXĐ : \(\hept{\begin{cases}\sqrt{x}+7\ne0\\2x+1\ne0\end{cases}}\)

<=> \(\hept{\begin{cases}\forall x\ge0\left(\text{vì }\sqrt{x}+7\ge7>0\right)\\x\ne-\frac{1}{2}\end{cases}}\)

<=> \(x\ge0\) 

\(a+b+\frac{4}{a}+\frac{1}{4b}=\frac{25}{36}a+\frac{4}{a}+\frac{25}{36}b+\frac{1}{4b}+\frac{11}{36}\left(a+b\right)\)

\(\ge2\sqrt{\frac{25}{36}a.\frac{4}{a}}+2\sqrt{\frac{25}{36}b.\frac{1}{4b}}+\frac{11}{36}.3\)

\(=\frac{10}{3}+\frac{5}{6}+\frac{11}{12}=\frac{61}{12}\)

Dấu \(=\)khi \(\hept{\begin{cases}\frac{25}{36}a=\frac{4}{a}\\\frac{25}{36}b=\frac{1}{4b}\\a+b=3\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{12}{5}\\b=\frac{3}{5}\end{cases}}\).

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}=\frac{1+a^2}{1+a^2}-\frac{a^2}{1+a^2}+\frac{1+b^2}{1+b^2}-\frac{b^2}{1+b^2}+\frac{1+c^2}{1+c^2}-\frac{c^2}{1+c^2}\)

\(=3-\left(\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)\)

\(\ge3-\left(\frac{a^2}{2a}+\frac{b^2}{2b}+\frac{c^2}{2c}\right)\)

\(=3-\frac{a+b+c}{2}=\frac{3}{2}\)

Dấu \(=\)khi \(a=b=c=1\).