1) A=\(\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}\) CM : A < \(\frac{504}{1009}\)
2) Cho a+c=2b; 2bd=c.(b-d) ( b,d \(\ne\) 0) CM \(\frac{a}{b}=\frac{c}{d}\)
3) Tìm x, y
\(\left(x-\frac{2}{5}\right)^{2018}+|y-2|=0\)
Những câu hỏi liên quan
so sánh 2 số A và B nếu
\(A=-\frac{1}{2018}-\frac{3}{2017^2}-\frac{5}{2017^3}-\frac{7}{2017^4};B=\frac{-1}{2018}-\frac{7}{2017^2}-\frac{5}{2017^3}-\frac{3}{2017^4}\)
So sánh A và B nếu
\(A=\frac{-1}{2018}-\frac{3}{2017^2}-\frac{5}{2017^3}-\frac{7}{2017^4}\)
\(B=\frac{-1}{2018}-\frac{7}{2017^2}-\frac{5}{2017^3}-\frac{3}{2017^4}\)
Cho A = \(\left(\frac{2}{3}\right)^2+\left(\frac{2}{5}\right)^2+\left(\frac{2}{7}\right)^2+......+\left(\frac{2}{2017}\right)^2\)
CM A < \(\frac{504}{1009}\)
thôi, không phải trả lời nữa, tui viết sai đề rồi còn đâu
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Uk , bt khai ra là tốt .
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a) Cho A= \(\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}\) CM: A< \(\frac{504}{1009}\)
b) Cho a+c= 2b và 2bd=c(b+d) (b, d không bằng 0). CM: \(\frac{a}{b}=\frac{c}{d}\)
\(a+c=2b\)
\(\Rightarrow2bd=\left(a+c\right).d=cb+cd\)
\(\Rightarrow ad+cd=cb+cd\)
\(\Rightarrow ad+cd-cd=cb\)
\(ad=cb\Rightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
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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(2^2.3right)^6+8^4.3^5}-frac{5^{10}.7^3-25^5.49}{left(125.7right)^3+5^9.14^3}C frac{left(a^{2016}+b^{2016}right)^{2017}}{left(c^{2016}+d^{2016}right)^{2017}} frac{left(a^{2017}-b^{2017}right)^{2016}}{left(c^{2017}-d^{2017}right)^{2016}}
Đọc tiếp
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 = \(\frac{\left(a^{2016}+b^{2016}\right)^{2017}}{\left(c^{2016}+d^{2016}\right)^{2017}}\)= \(\frac{\left(a^{2017}-b^{2017}\right)^{2016}}{\left(c^{2017}-d^{2017}\right)^{2016}}\)
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
cs ng làm đung r
đag định lm
Thực hiện phép tính:a, A frac{1}{6}.left(-2frac{3}{5}right)+1frac{2}{3}.left(frac{-13}{5}right)b, Bfrac{5}{7}:left(frac{1}{2}-frac{1}{3}right)+frac{5}{7}:left(frac{1}{5}-frac{1}{6}right)c, C 2^3+3.left(frac{2017}{2018}right)^0.left(frac{1}{2}right)^{-2}:4+[left(-2right)^2:frac{1}{2}].left(frac{-1}{2}right)^3giúp vs, mik cần gấp
Đọc tiếp
Thực hiện phép tính:
a, A= \(\frac{1}{6}.\left(-2\frac{3}{5}\right)+1\frac{2}{3}.\left(\frac{-13}{5}\right)\)
b, B=\(\frac{5}{7}:\left(\frac{1}{2}-\frac{1}{3}\right)+\frac{5}{7}:\left(\frac{1}{5}-\frac{1}{6}\right)\)
c, C= \(2^3+3.\left(\frac{2017}{2018}\right)^0.\left(\frac{1}{2}\right)^{-2}:4+[\left(-2\right)^2:\frac{1}{2}].\left(\frac{-1}{2}\right)^3\)
giúp vs, mik cần gấp
A= E387E4837
B = 883433
C = UỲUWFHQWURY48E3947
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CMR : \(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}< \frac{504}{1009}\)
chịu mà dã học đâu mà làm
làm theo lời đi
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Xem thêm câu trả lời
Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2017^2}\)
\(B=\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{2017!}\)
Chứng minh \(A+B< 2\)
A=1/2^2 + 1/3^2 + 1/4^2 + ... + 1/2017^2
A < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2016.2017
A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2016 - 1/2017
A < 1 - 1/2017 < 1 (1)
B = 2!/3! + 2!/4! + 2!/5! + ... + 2!/2017!
B = 2!.(1/3! + 1/4! + 1/5! + ... + 1/2017!)
B < 2.(1/2.3 + 1/3.4 + 1/4.5 + ... + 1/2016.2017)
B < 2.(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/2016 - 1/2017)
B < 2.(1/2 - 1/2017) < 2.1/2 = 1 (2)
Từ (1) và (2) => A + B < 2 (đpcm)
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cho \(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}.\)Chứng minh rằng :\(A< \frac{504}{1009}\)
\(A=2\cdot\left(\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2017^2}\right)< 2\cdot\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}\right)\)
Đặt \(M=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{2015\cdot2016}=\left(1+\frac{1}{3}+...+\frac{1}{2015}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
\(\Rightarrow M=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1008}\right)\)
\(\Rightarrow M=\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2016}< \frac{1}{1009}+\frac{1}{1009}+...+\frac{1}{1009}\)(1008 số hạng )
hay\(M< \frac{1008}{1009}\Rightarrow A< 2\cdot\frac{1008}{1009}=\frac{504}{1009}\left(ĐPCM\right)\)
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