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Đặt $x^2=u ;y^2=t$ Từ giả thiết : $\frac{u+t}{u-t}$ + $\frac{u+t}{u-t}$ = $a$$<=>$ $t^2= (\frac{a-2}{a+2})u^2$$<=>$ $A= \frac{x^8+y^8}{x^8-y^8}+ \frac{x^8-y^8}{x^8+y^8}= \frac{u^4+t^4}{u^4-t^4} + \frac{u^4-t^4}{u^4+t^4}= 2\frac{u^8+t^8}{u^8-t^8} = 2.\frac{u^8+(\frac{a-2}{a+2})^4u^8 }{u^8-(\frac{a-2}{a+2})^4u^8}=2.\frac{1+(\frac{a-2}{a+2})^4 }{1-(\frac{a-2}{a+2})^4}$ =..................

Đặt $x^2=u ;y^2=t$ Từ giả thiết : $\frac{u+t}{u-t}$ + $\frac{u+t}{u-t}$ = $a$$<=>$ $t^2= (\frac{a-2}{a+2})u^2$$<=>$ $A= $\frac{x^8+y^8}{x^8-y^8}+ \frac{x^8-y^8}{x^8+y^8}= \frac{u^4+t^4}{u^4-t^4} + \frac{u^4-t^4}{u^4+t^4}= 2\frac{u^8+t^8}{u^8-t^8} = 2.\frac{u^8+(\frac{a-2}{a+2})^4u^8 }{u^8-(\frac{a-2}{a+2})^4u^8}=2.\frac{1+(\frac{a-2}{a+2})^4 }{1-(\frac{a-2}{a+2})^4}$ =..................

Đặt $x^2=u ;y^2=t$ Từ giả thiết : $\frac{u+t}{u-t}$ + $\frac{u+t}{u-t}$ = $a$$<=>$ $t^2= (\frac{a-2}{a+2})u^2$$<=>$ $A= $\frac{x^8+y^8}{x^8-y^8}$ + $\frac{x^8-y^8}{x^8+y^8}$= $\frac{u^4+t^4}{u^4-t^4}$ + $\frac{u^4-t^4}{u^4+t^4}$= $2\frac{u^8+t^8}{u^8-t^8}$ = 2.$\frac{u^8+(\frac{a-2}{a+2})^4u^8 }{u^8-(\frac{a-2}{a+2})^4u^8}$=2.$\frac{1+(\frac{a-2}{a+2})^4 }{1-(\frac{a-2}{a+2})^4}$=...$Có gì mai sửagiờ đi ngủ đã :3...............

Đặt $x^2=u ;y^2=t$ Từ giả thiết : $\frac{u+t}{u-t}$ + $\frac{u+t}{u-t}$ = $a$$<=>$ $t^2= (\frac{a-2}{a+2})u^2$$<=>$ $A= \frac{x^8+y^8}{x^8-y^8}+ \frac{x^8-y^8}{x^8+y^8}= \frac{u^4+t^4}{u^4-t^4} + \frac{u^4-t^4}{u^4+t^4}= 2\frac{u^8+t^8}{u^8-t^8} = 2.\frac{u^8+(\frac{a-2}{a+2})^4u^8 }{u^8-(\frac{a-2}{a+2})^4u^8}=2.\frac{1+(\frac{a-2}{a+2})^4 }{1-(\frac{a-2}{a+2})^4}$ =..................

Đặt $x^2=u ;y^2=t$ Từ giả thiết : $\frac{u+t}{u-t}$ + $\frac{u+t}{u-t}$ = $a$$<=>$ $t^2= (\frac{a-2}{a+2})u^2$$<=>$ $A= $\frac{x^8+y^8}{x^8-y^8}$ + $\frac{x^8-y^8}{x^8+y^8}$= $\frac{u^4+t^4}{u^4-t^4}$ + $\frac{u^4-t^4}{u^4+t^4}$= $2\frac{u^8+t^8}{u^8-t^8}$ = 2.$\frac{u^8+(\frac{a-2}{a+2})^4u^8 }{u^8-(\frac{a-2}{a+2})^4u^8}$=2.$\frac{1+(\frac{a-2}{a+2})^4 }{1-(\frac{a-2}{a+2})^4}$=...$Mod chỉnh sửa hộ em :v Em đi ngủ đã :3...............

Đặt $x^2=u ;y^2=t$ Từ giả thiết : $\frac{u+t}{u-t}$ + $\frac{u+t}{u-t}$ = $a$$<=>$ $t^2= (\frac{a-2}{a+2})u^2$$<=>$ $A= $\frac{x^8+y^8}{x^8-y^8}$ + $\frac{x^8-y^8}{x^8+y^8}$= $\frac{u^4+t^4}{u^4-t^4}$ + $\frac{u^4-t^4}{u^4+t^4}$= $2\frac{u^8+t^8}{u^8-t^8}$ = 2.$\frac{u^8+(\frac{a-2}{a+2})^4u^8 }{u^8-(\frac{a-2}{a+2})^4u^8}$=2.$\frac{1+(\frac{a-2}{a+2})^4 }{1-(\frac{a-2}{a+2})^4}$=...$Có gì mai sửagiờ đi ngủ đã :3...............