So sánh A và B biết:\(a.A=\)\(-\frac{1}{2016}-\frac{3}{^{11^2}}-\frac{5}{11^3}-\frac{7}{11^4}\)\(B=-\frac{7}{11^2}-\frac{1}{2016}-\frac{5}{11^3}-\frac{3}{11^4}\)\(b.A=\frac{2006}{2007}-\frac{2007}{200...

\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)

\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)

\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

=> A > B

Vậy A > B

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HT

Hồ Thu Giang

25 tháng 8 2015

So sánh A và B biết:

A = \(-\frac{1}{2013}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)

B = \(-\frac{1}{2013}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)

#Toán lớp 6

1

TT

Trần Thị Loan

25 tháng 8 2015

\(A=\left(-\frac{1}{2013}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\right)-\frac{4}{11^4};B=\left(-\frac{1}{2013}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{3}{11^2}\right)-\frac{4}{11^2}\)

Vì 11⁴ > 11²nên \(\frac{4}{11^4}-\frac{4}{11^2}\) => A > B

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TT

thien ty tfboys

23 tháng 6 2015

So sánh A và B nếu :

\(A=-\frac{1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)va \(B=\frac{1}{2022}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)

#Toán lớp 7

0

MN

Mai Ngọc

17 tháng 9 2015

So sánh A và B

\(A=-\frac{1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)

\(B=-\frac{1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)

#Toán lớp 7

1

T

🤣🤣🤣 Ŧùɔ

1 tháng 7 2021

\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)

\(\text{B = }\text{ }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)

\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

=> A > B

Vậy A > B

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PM

phạm minh quang

11 tháng 9 2018

Bài 1 ; So sánh

\(A=\frac{-1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)

\(B=\frac{1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)

Mình cần gấp lắm ạ , Ai làm đúng và nhanh nhất mình tick cho

#Toán lớp 7

1

T

🤣🤣🤣 Ŧùɔ

1 tháng 7 2021

\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)

\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)

\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

=> A > B

Vậy A > B

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NN

Nhóc Nhí Nhảnh

17 tháng 7 2016

Bài 1: So sánh

a) A=\(\frac{-1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)

B=\(\frac{-1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)

#Toán lớp 7

0

NL

ngọc linh

26 tháng 10 2019

A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49}{\left(125.7\right)^3+5^9.14^3}\)C...

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#Toán lớp 7

3

XO

Xyz OLM

26 tháng 10 2019

A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)

\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)

\(=\frac{3}{5}+\frac{2}{5}=1\)

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XO

Xyz OLM

26 tháng 10 2019

b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)

\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)

\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)

\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)

\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)

\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)

\(=\frac{7}{42}-\frac{20}{42}\)

\(=-\frac{13}{42}\)

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H

huy

16 tháng 8 2017

So sánh A và B:A=- -\(\frac{1}{2011}\)-\(\frac{3}{11^2}\)-\(\frac{5}{11^3}\)-\(\frac{7}{11^4}\)

B= --\(\frac{1}{2011}\)-\(\frac{7}{11^2}\)-\(\frac{5}{11^3}\)-\(\frac{3}{11^4}\)

#Toán lớp 7

0

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