Question

Nếu biết \(\frac{sin^4a}{a}+\frac{cos^4a}{b}=\frac{1}{a+b}\) thì biểu thức \(M=\frac{sin^{10}a}{a^4}+\frac{cos^{10}a}{b^4}\) bằng

A. \(\frac{1}{\left(a+b\right)^5}\)

B. \(\frac{1}{a^5}+\frac{1}{b^5}\)

C. \(\frac{1}{a^4}+\frac{1}{b^4}\)

D. \(\frac{1}{\left(a+b\right)^4}\)

Answer 1

\(\frac{\sin^4\alpha}{a}+\frac{\cos^4\alpha}{b}\ge\frac{\left(\sin^2\alpha+\cos^2\alpha\right)^2}{a+b}=\frac{1}{a+b}\)

\("="\Leftrightarrow\frac{\sin^2\alpha}{a}=\frac{\cos^2\alpha}{b}\Leftrightarrow\sin^2\alpha.b=a-a.\sin^2\alpha\)

\(\Leftrightarrow\sin^2\alpha\left(b+a\right)=a\Rightarrow\sin^2\alpha=\frac{a}{a+b}\)

\(\cos^2\alpha.a=b-b\cos^2\alpha\Rightarrow\cos^2\alpha=\frac{b}{a+b}\)

\(\Rightarrow M=\frac{\frac{a^5}{\left(a+b\right)^5}}{a^4}+\frac{\frac{b^5}{\left(a+b\right)^5}}{b^4}=\frac{a+b}{\left(a+b\right)^5}=\frac{1}{\left(a+b\right)^4}\) => D