\(B=1+\dfrac{1}{13}+\dfrac{1}{13^2}+...+\dfrac{1}{13^n}\)

\(=>13B=13+1+\dfrac{1}{13}+...+\dfrac{1}{13^{n-1}}\)

\(=>13B-B=13-\dfrac{1}{13^n}\)

\(=>12B=\dfrac{13^{n+1}-1}{13^n}\)

\(=>B=\dfrac{13^{n+1}-1}{13^n.12}\)

\(#Nttnam\)

\(\dfrac{\sqrt{13}}{\sqrt{\sqrt{13}+1-1}}-\dfrac{\sqrt{13}}{\sqrt{\sqrt{13}+1+1}}\\ =\dfrac{\sqrt{13}}{\sqrt{\sqrt{13}}}-\dfrac{\sqrt{13}}{\sqrt{\sqrt{13}+2}}\\ =\sqrt{\dfrac{13}{\sqrt{13}}}-\sqrt{\dfrac{13}{\sqrt{13}+2}}\\ =\sqrt[4]{13}-\sqrt{\dfrac{13\sqrt{13}-26}{9}}\\=\sqrt[4]{13}-\dfrac{\sqrt{13\sqrt{13}-26} }{3}\\ =\dfrac{3\sqrt[4]{13}-\sqrt{13\sqrt{13}-26}}{3}\)

\(ta có A=\dfrac{13^{15}+1}{13^{16}+1}=\dfrac{13^{15}}{13^{16}}+1\)=\(\dfrac{1}{13}+1\)

B=\(\dfrac{13^{16}+1}{13^{17}+1}=\dfrac{13^{16}}{13^{17}}+1\)=\(\dfrac{1}{13}+1\)

vậy A=B

\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)

ta có

\(\dfrac{13^{16}+1}{13^{17}+1}< 1\Rightarrow\dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)

vậy B<A

\(A=\dfrac{13^{15}+1}{13^{16}+1}vàB=\dfrac{13^{16}+1}{13^{17}+1}\)

ta có B<1 nên

\(\dfrac{13^{16}+1}{13^{17}+1}< \dfrac{13^{16}+1+12}{13^{17}+1+12}=\dfrac{13\left(13^{15}+1\right)}{13\left(13^{16}+1\right)}=\dfrac{13^{15}+1}{13^{16}+1}=A\)

Vậy B<A

cacwj con

1/13.(8/13-5/13)+14/13

1/13.3/13+14/13

3/13+14/13

17/13

NHƯ VẦY ĐÓ BẠN MIK CŨNG HỌC LỚP 6 NHƯ BẠN ĐÓ

dấu chấm là gì vậy bạn
 

dấu nhân mà !!!!

1/13 . 8/13 + 5/13 . 1/13 - 14/13

= 1/13 . (8/13 + 5/13) - 14/13

= 1/13 . 13/13 - 14/13

= 1/13 . 1 - 14/13

= 1/13 - 14/13

= -13/13 

= -1