Cho $\frac{sin^4x}{a} + \frac{cos^4x}{b} = \frac{1}{a+b}$
Chứng minh rằng: $\frac{sin^8x}{a^3}+\frac{cos^8x}{b^3}=\frac{1}{(a+b)^3}$
Ta có $ \dfrac{1}{a+b}= \dfrac{\sin^{4}x}{a}+\dfrac{\cos^{4}x}{b} \ge \dfrac{(\sin^2x+\cos^2x)^2}{a+b}=\dfrac{1}{a+b}$
Dâu bằng xảy ra khi: $\dfrac{\sin^2x}{a}=\dfrac{\cos^2x}{b}\Rightarrow \dfrac{\sin^2x}{a}=\dfrac{1-\sin^2x}{b} \Rightarrow (a+b)\sin^2x=a \Rightarrow \sin^2x=\frac{a}{a+b}$. 
Tương tự: $\cos^2x=\frac{b}{a+b}$.
Suy ra: $ \frac{\sin^{8}x}{a^3}+\frac{\cos^{8}x}{b^{3}}=\frac{1}{(a+b)^3}$

Ta coi cái đề cho là (1') và cái đề cần CM là (2'):
Thì từ (1) ; bạn quy đồng hết lên và sử dụng công thức : $sin^4x+cos^4x=1-\frac{1}{2}sin^2 2x$
Từ đó ta thu được 1 biểu thức là : $(acos^2x-bsin^2x)^2=0 \Leftrightarrow acos^2x+bsin^2x=0$     (2)
Rút từ (2) ta có : $sin^2x=\frac{a}{a+b} \Rightarrow cos^2x=\frac{b}{a+b}$ (3)
Còn (2') ta phân tích : $sin^8x=(sin^2x)^4 và cos^8x=(cos^2x)^4 $ 
Lầy (3) thế vào (2') thì dược điều ta cần CM

Bạn cần đăng nhập để có thể gửi đáp án

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