$\frac{x^2-y^2}{x^2+y^2}$+$\frac{x^2+y^2}{x^2+y^2}$=a. Tính giá trị biểu thức $M= \frac{x^8+y^8}{x^8-y^8}$+$\frac{x^8-y^8}{x^8+y^8}$ theo a
Ta có $\frac{x^2-y^2}{x^2+y^2}$+$\frac{x^2+y^2}{x^2-y^2}$=a
$\Rightarrow \frac{2(x^4+y^4)}{(x^2+y^2)(x^2-y^2)}$=a
$\Rightarrow \frac{2(x^4+y^4)}{x^4-y^4}$=a
$\Rightarrow 2(x^4+y^4)=(x^4-y^4)a$
$\Rightarrow y^4(a+2)=x^4(a-2)$
$\Rightarrow \frac{y^4}{x^4}=\frac{a-2}{a+2}$
$\Rightarrow \frac{y^{16}}{x^{16}}=(\frac{a-2}{a+2})^4$
$\Rightarrow \frac{y^{16}}{x^{16}}+1=(\frac{a-2}{a+2})^4+1$
$\Rightarrow \frac{y^{16}+x^{16}}{x^{16}}=(\frac{a-2}{a+2})^4+1$ (1)
Lại có:$\frac{y^4}{x^4}=\frac{a-2}{a+2}$
$\Rightarrow \frac{y^{16}}{x^{16}}=(\frac{a-2}{a+2})^4$
$\Rightarrow \frac{y^{16}}{x^{16}}=(\frac{a-2}{a+2})^4$
$\Rightarrow \frac{y^{16}}{x^{16}}-1=(\frac{a-2}{a+2})^4-1$
$\Rightarrow \frac{y^{16}-x^{16}}{x^{16}}=(\frac{a-2}{a+2})^4-1$
$\Rightarrow \frac{x^{16}-y^{16}}{x^{16}}=1-(\frac{a-2}{a+2})^4$ (2)
Lấy (1) chia (2)
$\Rightarrow \frac{\frac{y^{16}+x^{16}}{x^{16}}}{\frac{x^{16}-y^{16}}{x^{16}}}$= $\frac{(\frac{a-2}{a+2})^4+1}{1-(\frac{a-2}{a+2})^4}$
$\Rightarrow \frac{y^{16}+x^{16}}{x^{16}-y^{16}}$= $\frac{(\frac{a-2}{a+2})^4+1}{1-(\frac{a-2}{a+2})^4}$
$\Rightarrow \frac{2y^{16}+2x^{16}}{x^{16}-y^{16}}$= 2$\frac{(\frac{a-2}{a+2})^4+1}{1-(\frac{a-2}{a+2})^4}$
$\Rightarrow \frac{y^{16}+x^{16}+2y^8x^8+y^{16}+x^{16}-2y^8x^8}{x^{16}-y^{16}}$= 2$\frac{(\frac{a-2}{a+2})^4+1}{1-(\frac{a-2}{a+2})^4}$
$\Rightarrow \frac{(x^8-y^8)^2+(x^8+y^8)^2}{(x^8-y^8)(x^8+y^8)}$= 2$\frac{(\frac{a-2}{a+2})^4+1}{1-(\frac{a-2}{a+2})^4}$
$\Rightarrow \frac{x^8-y^8}{(x^8+y^8)}+\frac{x^8+y^8}{(x^8-y^8)}$= 2$\frac{(\frac{a-2}{a+2})^4+1}{1-(\frac{a-2}{a+2})^4}$,
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
