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21 tháng 8 2023
\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)
\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)
\(X+16=-5\)
\(X=-21\)
20 tháng 4 2017
\(\frac{4}{3.5}+\frac{8}{5.9}+\frac{12}{9.15}+...+\frac{32}{n\left(n+16\right)}=\frac{16}{25}\)
\(2\left(\frac{1}{3}-\frac{1}{5}\right)+2\left(\frac{1}{5}-\frac{1}{9}\right)+2\left(\frac{1}{9}-\frac{1}{15}\right)+...+2\left(\frac{1}{n}-\frac{1}{n+16}\right)=\frac{16}{25}\)
\(2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+16}\right)=\frac{16}{25}\)
\(2\left(\frac{1}{3}-\frac{1}{n+16}\right)=\frac{16}{25}\)
\(\frac{1}{3}-\frac{1}{n+16}=\frac{8}{25}\)
\(\frac{1}{n+16}=\frac{1}{75}\)
\(\Rightarrow n+16=75\)
\(\Rightarrow n=59\)
HL
1
BT
22 tháng 3 2018
\(M=\frac{1}{3.4}+\frac{7}{3.4}+\frac{2}{3.5}+\frac{14}{5.9}-\frac{4}{9.13}=\frac{8}{3.4}+\frac{2}{3.5}+\frac{2}{9}\left(\frac{7}{5}-\frac{2}{13}\right)\)
=> \(M=\frac{2}{3}+\frac{2}{3.5}+\frac{2}{9}.\frac{81}{5.13}=\frac{2}{3}\left(1+\frac{1}{5}\right)+\frac{18}{5.13}\)
=> \(M=\frac{2}{3}.\frac{6}{5}+\frac{18}{5.13}=\frac{4}{5}+\frac{18}{5.13}=\frac{2}{5}\left(2+\frac{9}{13}\right)=\frac{2}{5}.\frac{35}{13}\)
=> \(M=\frac{14}{13}\)
21 tháng 2 2022
\(\dfrac{-8}{15}.\dfrac{3}{7}.\dfrac{15}{-8}.14=\left(\dfrac{-8}{15}.\dfrac{15}{-8}\right).\left(\dfrac{3}{7}.14\right)=1.6=6\)
\(\dfrac{17}{27}.\dfrac{13}{9}-\dfrac{4}{9}.\dfrac{17}{27}=\dfrac{17}{27}.\left(\dfrac{13}{9}-\dfrac{4}{9}\right)=\dfrac{17}{27}.1=\dfrac{17}{27}\)
KT
1
TN
14 tháng 4 2016
bài 1
\(2A=\left(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{99\cdot101}\right)\cdot2\)
\(=5\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\right)\)
\(=5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=5\left(1-\frac{1}{101}\right)\)
\(=5\cdot\frac{100}{101}\)
\(=\frac{500}{101}\Rightarrow A=\frac{500}{101}:2=\frac{250}{101}\)
bài 2:
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=-\frac{37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(x+\frac{8}{45}=-\frac{37}{45}\)
\(x=-\frac{37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=-1\)
XO
4 tháng 8 2020
Ta có \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)(đk : \(x\ne0\))
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
=> \(\frac{7}{x}=\frac{7}{15}\)
=> x = 15 (tm)
b) \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)
=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
=> \(\frac{1}{3}-\frac{1}{n+3}=\frac{10}{31}\)
=> \(\frac{1}{2x+3}=\frac{1}{93}\)
=> 2x + 3 = 93
=> 2x = 90
=> x = 45
21 tháng 10 2018
ban nao co chuyen shin ko cho minh muon minh giai cho 10 bai nhu the i love pac pac
LN
28 tháng 1 2019
\(A=\frac{2^{12}.3^4-4^5.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)
\(A=\frac{2^{12}.3^4-2^{10}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(A=\frac{2^{10}.3^4\left(2^2-1\right)}{2^{10}.3^4\left(2^2.3^2+2^2.3\right)}\)
\(A=\frac{2^2-1}{2^2.3^2+2^2.3}\)
\(A=\frac{4-1}{36+12}\)
\(A=\frac{3}{48}=\frac{1}{16}\)
Không chép lại đầu bài nha!
=\(\frac{5-3}{3.5}+\frac{9-5}{5.9}+\frac{17-9}{9.17}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(\frac{1}{3}-\frac{1}{11}\)
=\(\frac{8}{33}\)