Ta có A= -1+7+(-7²)+7³+(-7⁴)+....+7²⁰⁰⁸ +7²⁰⁰⁸

        A.7=[-7+7²+(-7³⁾+7⁴+....+7²⁰⁰⁹ +7^{2009]  + [} -1+7+(-7²)+7³+(-7⁴)+....+7²⁰⁰⁸ +7^2008]

        A.7=[7²⁰⁰⁹.2+(-1) +7^2008] :7

b;c làm tương tự

\(\frac{7+\sqrt{7}}{1+\sqrt{7}}+\frac{7-\sqrt{7}}{1-\sqrt{7}}=\frac{\sqrt{7}\left(1+\sqrt{7}\right)}{1+\sqrt{7}}+\frac{\sqrt{7}\left(1-\sqrt{7}\right)}{1-\sqrt{7}}=\sqrt{7}+\sqrt{7}=2\sqrt{7}\)

\(\left(\frac{\sqrt{7}-7}{1-\sqrt{7}}-\frac{\sqrt{7}+1}{7+\sqrt{7}}\right):\frac{\sqrt{7}+1}{\sqrt{7}}-7\sqrt{\frac{1}{\sqrt{7}}}=\left(\sqrt{7}-\sqrt{7}\right):\frac{\sqrt{7}+1}{\sqrt{7}}-7\sqrt{\frac{1}{\sqrt{7}}}=-7\sqrt{\frac{1}{\sqrt{7}}}\)

C/M : A<\(\frac{1}{50}\)

Ta có

A = \(\dfrac{1+7+7^2+7^3+...+7^{11}}{1+7+7^2+7^3+...+7^{10}}\)

Đặt C = 1 + 7 + 7² + 7³+...+7¹¹

7C = 7 + 7² + 7³ + ... + 7¹¹ + 7¹²

=> 6C = 7¹² - 1

C = \(\dfrac{7^{12}-1}{6}\)

Đặt D = 1 + 7 + 7² + 7³+...+7¹⁰

7D = 7 + 7² + 7³ + ... + 7¹⁰ + 7¹¹

=> 6D = \(7^{11}-1\)

D = \(\dfrac{7^{11}-1}{6}\)

=> A = \(\dfrac{\dfrac{7^{12}-1}{6}}{\dfrac{7^{11}-1}{6}}\)

A = \(\dfrac{7^{12}-1}{6}\) : \(\dfrac{7^{11}-1}{6}\)

A = \(\dfrac{7^{12}-1}{6}.\dfrac{6}{7^{11}-1}\)

A = \(\dfrac{7^{12}-1}{7^{11}-1}\) = 7, 000000003

Lại có:

B = \(\dfrac{1+3+3^2+3^3+...+3^{11}}{1+3+3^2+3^3+...+3^{10}}\)\

Đặt H = \(1+3+3^2+3^3+...+3^{11}\)

3H = \(3+3^2+3^3+...+3^{12}\)

=> 2H = \(3^{12}-1\)

H = \(\dfrac{3^{12}-1}{2}\)

Đặt Q = \(1+3+3^2+3^3+...+3^{10}\)

3Q = \(3+3^2+3^3+...+3^{10}+3^{11}\)

=> 2Q = \(3^{11}-1\)

Q = \(\dfrac{3^{11}-1}{2}\)

=> B = \(\dfrac{\dfrac{3^{12}-1}{2}}{\dfrac{3^{11}-1}{2}}\)

B = \(\dfrac{3^{12}-1}{2}:\dfrac{3^{11}-1}{2}\)

B = \(\dfrac{3^{12}-1}{2}.\dfrac{2}{3^{11}-1}\)

B = \(\dfrac{3^{12}-1}{3^{11}-1}\)

B = 3, 00001129

Vì 7, 000000003 > 3, 00001129

=> A > B

Vậy A > B

\(\left(\dfrac{1}{7}\right)^7.7^7=\dfrac{1^7}{7^7}.7^7=1^7=1\)

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 \(A< 3\)

b) \(A=\frac{1+5+5^2+5^3+...+5^{10}+5^{11}}{1+5+5^2+5^3+...+5^9+5^{10}}=5^{11}\)

bn rút gọn là dc  

\(B=\frac{1+7+7^2+7^3+...+7^{10}+7^{11}}{1+7+7^2+7^3+...+7^9+7^{10}}=7^{11}\)

\(A=5^{11},B=7^{11}\)

\(\Rightarrow7^{11}>5^{11}\Rightarrow B>A\)

hk tốt #