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19 tháng 11 2019

Bạn tham khảo tại đây nhé: Câu hỏi của Nguyễn Chí Thành.

Chúc bạn học tốt!

19 tháng 11 2019

\(A=\frac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}=\frac{1+5\left(1+5+5^2+...+5^9\right)}{1+5+5^2+...+5^9}=\frac{1}{1+5+5^2+...+5^9}+5\)

\(B=\frac{1+3+...+3^{10}}{1+3+...+3^9}=\frac{1+3\left(1+3+...+3^9\right)}{1+3+...+3^9}=\frac{1}{1+3+...+3^9}+3\)

\(\Rightarrow A>B\)

a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)

ta có :

 \(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)

 \(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)

\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)

Vậy \(A< 3\)

2 tháng 5 2019

a. Ta có :

\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)

\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)

\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)

Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)

Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)

Vậy \(A< 3\)

20 tháng 5 2018

a) Đặt \(A=\frac{7^{15}}{1+7+7^2+...+7^{14}}\)

Đặt \(B=1+7+7^2+...+7^{14}\)

\(\Rightarrow7B=7+7^2+...+7^{15}\)

\(\Rightarrow7B-B=6B=7^{15}-1\)

\(\Rightarrow B=\frac{7^{15}-1}{6}\)

\(\Rightarrow A=\frac{7^{15}-1+1}{\frac{7^{15}-1}{6}}=\left(7^{15}-1\right).\frac{6}{7^{15}-1}+\frac{6}{7^{15}-1}=6+\frac{6}{7^{15}-1}\)

Tự làm tiếp nha

21 tháng 5 2018

bạn giải nốt đi

2 tháng 1 2018

\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)

\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)

\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)  

Có: \(\frac{1}{1+5+5^2+...+5^8}>0\)              và      \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)

\(\Rightarrow A>B\)

25 tháng 3 2019

ta có : A = \(\frac{7^{10}}{1+7+7^2+7^3+...+7^9}=1:\frac{1+7+7^2+7^3+...+7^9}{7^{10}}\)

\(1:\left(\frac{1}{7^{10}}+\frac{7}{7^{10}}+\frac{7^2}{7^{10}}+...+\frac{7^8}{7^{10}}+\frac{7^9}{7^{10}}\right)\)=\(1:\left(\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7^2}+\frac{1}{7}\right)\)

tương tự ta được : B = \(1:\left(\frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5^2}+\frac{1}{5}\right)\)

Vì \(\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7^2}+\frac{1}{7}\)\(\frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5^2}+\frac{1}{5}\)

=> A > B 

7 tháng 7 2019

a, \(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)

b, Ta có: \(\frac{1}{A}=\frac{2^{20}-3}{2^{18}-3}=\frac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\frac{9}{2^{18}-3}\)

\(\frac{1}{B}=\frac{2^{22}-3}{2^{20}-3}=\frac{2^2\left(2^{20}-3\right)+9}{2^{20}-3}=4+\frac{9}{2^{20}-3}\)

Vì \(\frac{9}{2^{18}-3}>\frac{9}{2^{20}-3}\)\(\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)

c,  Câu hỏi của truong nguyen kim 

12 tháng 6 2018

b, Ta có:\(\dfrac{1+3+3^2+.....+3^{10}}{1+3+3^2+.....+3^9}\) \(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3+3^2+...+3^{10}}{1+3+3^2+...+3^9}\)\(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3.\left(1+3+3^2+...+3^9\right)}{1+3+3^2+...+3^9}\)

\(=\dfrac{1}{1+3+3^2+...+3^9}+3< 4\)

\(\Rightarrow\) \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< 4\) \(\left(1\right)\)

Ta có :\(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5+5^2+...+5^{10}}{1+5+5^2+....+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5.\left(1+5+5^2+...+5^9\right)}{1+5+5^2+...+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+5>5\)

\(\Rightarrow\) \(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}>5\) \(\left(2\right)\)

Từ \(\left(1\right)và\left(2\right)\)

\(\Rightarrow\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

Vậy \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

12 tháng 6 2018

a, Đặt \(A\)\(=\dfrac{7^{15}}{1+7+7^2+...+7^{14}}\)

\(\Rightarrow\) \(\dfrac{1}{A}\) \(=\dfrac{1+7+7^2+...+7^{14}}{7^{15}}=\dfrac{1}{7^{15}}+\dfrac{7}{7^{15}}+\dfrac{7^2}{7^{15}}+...+\dfrac{7^{14}}{7^{15}}\)

\(=\dfrac{1}{7^{15}}+\dfrac{1}{7^{14}}+\dfrac{1}{7^{13}}+....+\dfrac{1}{7}\)

Đặt \(B=\dfrac{9^{15}}{1+9+9^2+...+9^{14}}\)

\(\Rightarrow\dfrac{1}{B}=\dfrac{1+9+9^2+...+9^{14}}{9^{15}}=\dfrac{1}{9^{15}}+\dfrac{9}{9^{15}}+\dfrac{9^2}{9^{15}}+...+\dfrac{9^{14}}{9^{15}}\)

\(=\dfrac{1}{9^{15}}+\dfrac{1}{9^{14}}+\dfrac{1}{9^{13}}+...+\dfrac{1}{9}\)

\(\dfrac{1}{7^{15}}>\dfrac{1}{9^{15}};\dfrac{1}{7^{14}}>\dfrac{1}{9^{14}};\dfrac{1}{7^{13}}>\dfrac{1}{9^{13}};....;\dfrac{1}{7}>\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{A}>\dfrac{1}{B}\) \(\Rightarrow A< B\)

Vậy\(\dfrac{7^{15}}{1+7+7^2+...+7^{14}}>\dfrac{9^{15}}{1+9+9^2+....+9^{14}}\)