Tìm a
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right).\)
Những câu hỏi liên quan
a)Tìm số nguyên dương n thỏa mãn:
\(\frac{1}{2}.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n.\left(n+2\right)}\right)=\frac{2013}{2014}\)
b)tìm a sao cho
\(\left(a+\frac{1}{1.3}\right)+\left(a+\frac{1}{3.5}\right)+\left(a+\frac{1}{5.7}\right)+...+\left(a+\frac{1}{23.25}\right)=11.a+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
Tĩm x:
1,\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+5}\)
2,\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}+......+x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
Giải chi tiết hộ mình nha mình tik cho
17/lim\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\right)\)
18/lim\(\frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}\left(\left|a\right|< 1;\left|b\right|< 1\right)\)
19/lim\(\frac{1-2+3-4+...+\left(2n-1\right)-2n}{2n+1}\)
\(A=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{3.5}\right).\left(1+\frac{1}{5.7}\right).....\left(1+\frac{1}{2015.2017}\right)\)
Tính A
Lời giải:
Xét tổng quát:
1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)1+1k(k+2)=k(k+2)+1k(k+2)=(k+1)2k(k+2)
Thay k=1,2,....,2015k=1,2,....,2015 ta có:
1+11.3=221.31+11.3=221.3
1+12.4=322.41+12.4=322.4
1+13.5=423.51+13.5=423.5
1+14.6=524.61+14.6=524.6
.............
1+12015.2017=201622015.20171+12015.2017=201622015.2017
Nhân theo vế:
⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)⇒A=12(1+11.3)(1+12.4)(1+13.5)....(1+12015.2017)
=12.221.3.322.4.423.5.524.6....201622015.2017=12.221.3.322.4.423.5.524.6....201622015.2017
=(1.2.3...2016)2(1.2.3...2015)(2.3.4...2017)=(1.2.3...2016)(2.3....2016)(1.2.3...2015)(2.3.4...2017)=2016.12017=20162017
Tính giá trị của biểu thức:Tfrac{4}{2.4}+frac{4}{4.6}+frac{4}{6.8}+...+frac{4}{2008.2010}Afrac{1}{1.3}+frac{1}{3.5}+frac{1}{5.7}+...+frac{1}{2007.2009}+frac{1}{2009.2011}Cleft(1-frac{1}{2}right)left(1-frac{1}{3}right)left(1-frac{1}{4}right)...left(1-frac{1}{1000}right)Sfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}Bleft(1+frac{1}{2}right)left(1+frac{1}{3}right)left(1+frac{1}{4}right)...left(1+frac{1}{100}right)Dleft(1-frac{1}{17}right)left(1-frac{2}{17}right)left(1-frac{3}{17}right)...left(1-f...
Đọc tiếp
Tính giá trị của biểu thức:
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(S=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{100}\right)\)
\(D=\left(1-\frac{1}{17}\right)\left(1-\frac{2}{17}\right)\left(1-\frac{3}{17}\right)...\left(1-\frac{27}{17}\right)\)
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
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\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
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\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(C=\frac{1.2.3...99}{2.3.4...100}\)
\(\Rightarrow C=\frac{1}{100}\)
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Xem thêm câu trả lời
A =\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\) B=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n-1\right).\left(2n+1\right)}\)
A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{3.7}\)\(=\frac{4}{7}\)
Tìm giá trị của biểu thức sau:
a)\(A=\frac{2}{3}+\frac{5}{6}:5-\frac{1}{18}.\left(-3\right)^2\)
b)\(B=3.\left\{5.\left[\left(5^2+2^3\right):11\right]-16\right\}+2015\)
c)\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2014.2016}\right)\)
https://dethihsg.com/de-thi-hoc-sinh-gioi-phong-gddt-hoang-hoa-2014-2015/
vào đây gợi ý nhé
k mik đi
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Xem thêm câu trả lời
tìm x nguyên biết
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{\left(2x-1\right)\left(2x+1\right)}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{\left(2x-1\right)}-\frac{1}{\left(2x+1\right)}\)
\(2A=1-\frac{1}{2x+1}=\frac{2x}{2x+1}\)
\(A=\frac{x}{2x+1}\)
Mà \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)
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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right)\left(2x-1\right)}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}+\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)
\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)
\(\Rightarrow99x=49\left(2x+1\right)\)
\(\Rightarrow99x=98x+49\)
\(\Rightarrow x=49\)
Vậy : \(x=49\)
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A = \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
B = 3.{ 5 . [ ( 5^2 + 2^3 ) : 11 ] -16} + 2015
C = \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right).....\left(1+\frac{1}{2014.2016}\right)\)