Đặt A= \(\frac{3^{10}+1}{3^9+1}\) đặt B= \(\frac{3^{11}+1}{3^{10}+1}\)

Vì B<1 => B< \(\frac{3^{11}+1+2}{3^{10}+1+2}\) = \(\frac{3^{11}+3}{3^{10}+3}\) = \(\frac{3\cdot\left(3^{10}+1\right)}{3\cdot\left(3^9+1\right)}\) =  \(\frac{3^{10}+1}{3^9+1}\) = A

Vậy B<A

Ta có :

\(\frac{3^{11}+1}{3^{10}+1}>1\) nên  \(\frac{3^{11}+1}{3^{10}+1}>\frac{3^{11}+1+2}{3^{10}+1+2}=\frac{3^{11}+3}{3^{10}+3}=\frac{3\left(3^{10}+1\right)}{3\left(3^9+1\right)}=\frac{3^{10}+1}{3^9+1}\)

Vậy \(\frac{3^{11}+1}{3^{10}+1}>\frac{3^{10}+1}{3^9+1}\)

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 \(A< 3\)

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 \(A< 3\)

TL :

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HT

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 Theo  thứ tự nhé

a) <

b) <

c) >

Đặt C =  A - 3 = \(1+3+3^2+...+3^{10}\)

  \(\Rightarrow3C=3+3^2+3^3+...+3^{11}\)

\(\Rightarrow2C=3C-C=3^{11}-1\)

\(\Rightarrow C=\frac{3^{11}-1}{2}\) \(\Rightarrow A-3=\frac{3^{11}-1}{2}=\frac{3^{11}}{2}-\frac{1}{2}\)

\(\Rightarrow A=\frac{3^{11}}{2}-\frac{1}{2}+3=\frac{3^{11}}{2}+\frac{5}{2}>\frac{3^{11}}{2}=B\)

Vậy A > B

A = 1 + 3 + 3² + 3³ + ....... + 3¹⁰ và B = 3¹¹ / 2

Ta có A = 1 + 3 + 3² +....+ 3¹⁰

3A = 3. ( 1 + 3 + .... + 3¹⁰ )

3A = 3 + 3² + 3³ +.......+ 3¹¹

3A - A = (3 + 3² + 3³+ ...+ 3¹¹)- ( 1 + 3 + ....+ 3¹⁰)

2A = 3¹¹ - 1

A = 3¹¹ - 1 / 2 thì < 3¹¹ / 2

=> A < B

Câu hỏi của Quỳnh Anh - Toán lớp 6 - Học toán với OnlineMath

Em tham khảo câu 1 2 cách 2 bạn hướng dẫn nhé!

Bài 1:

\(A=\frac{3333}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{3333}{101}\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(A=\frac{3333}{101}\left(\frac{1}{3}-\frac{1}{7}\right)=\frac{3333}{101}.\frac{4}{21}=\frac{1111.4}{101.7}=\frac{4444}{707}\)

Bài 2

\(A=\frac{2^{10}+1}{2^{10}-1}=\frac{2^{10}-1+2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)

\(B=\frac{2^{10}-1}{2^{10}-3}=\frac{2^{10}-3+4}{2^{10}-3}=1+\frac{4}{2^{10}-3}\)

Ta thấy \(2^{10}-1>2^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}< \frac{4}{2^{10}-3}\)

Từ đó \(\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{4}{2^{10}-3}\Rightarrow A< B\)

Bài 3\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}\right)+\frac{5}{11}}{\frac{5}{12}+\left(1-\frac{7}{11}\right)}=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}=\frac{\frac{55+60}{11.12}}{\frac{55+48}{12.11}}=\frac{115}{103}\)

Bài 2 sai r bạn ơi