Tính :
\(A=\frac{1}{3}-\frac{1}{18}-\frac{1}{54}-\frac{1}{108}-\frac{1}{180}-\frac{1}{270}-\frac{1}{378}\)
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27 tháng 6 2022
3H = 3 (1/3-1/8-1/54-1/108-1/180-1/270-1/378)
3H+(3/8) = 3/3 - 3/(6x9) - 3/(9x12) - 3/(12x15) - 3/(15x18) - 3/(1/18x21)
3H+(3/8) = 1 - [(1/6)-(1/9)] - [(1/9)-(1/12)] - [(1/12)-(1/15)] - [(1/15)-(1/18)] - [(1/18)-(1/21)]
= 1 -1/6 +1/9 -1/9 +1/12 -1/12 +1/15 -1/15 +1/18 -1/18 +1/21
3H = -(3/8) + 1 -(1/6) +(1/21)
3H = (-63 +168 -28 +8) / 168
3H = 85/168
đáp số:
H = 85/504
gần bằng 0,17
12 tháng 12 2018
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)(điều kiện: \(x\ne\left\{-4;-5;-6;-7\right\}\) )
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow54=\left(x+4\right)\left(x+7\right)\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-13\\x=2\end{cases}}\)(thỏa mãn ĐKXĐ)
Vậy tập nghiệm của pt là: \(S=\left\{-13;2\right\}\)
12 tháng 12 2018
Lâu lắm không làm nhể
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\frac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\frac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)
Dùng công thứ \(\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
Khi đó \(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{x+7}{\left(x+4\right).\left(x+7\right)}-\frac{\left(x+4\right)}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow\left(x+4\right).\left(x+7\right)=54\)
\(\Rightarrow\hept{\begin{cases}x+4=6\\x+7=9\end{cases}}\)hoặc \(\hept{\begin{cases}x+4=-6\\x+7=-9\end{cases}}\)
Suy ra \(x=3\)hoặc \(x=-3\)
0
3
2 tháng 8 2018
\(\frac{1}{3}-\left(\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+\frac{1}{180}+\frac{1}{270}+\frac{1}{378}\right)\)
\(=\frac{1}{3}-\left(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\frac{1}{12.15}+\frac{1}{15.18}+\frac{1}{18.21}\right)\)
\(=\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(=\frac{1}{3}-\frac{1}{3}+\frac{1}{21}=\frac{1}{21}\)
11 tháng 9 2018
từ đề bài ta có \(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)
DA
29 tháng 5 2020
A=\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\)
\(\frac{1}{2}\)A= \(\frac{1}{2}.\left(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2018.2020}\right)\)
\(\frac{1}{2}A\)= \(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2018.2020}\)
\(\frac{1}{2}A\)= \(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{2020-2018}{2018.2020}\)
\(\frac{1}{2}A\)= \(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2018}-\frac{1}{2020}\)
\(\frac{1}{2}A\)= \(\frac{1}{2}-\frac{1}{2020}\)
\(\frac{1}{2}A=\frac{1009}{2020}\)
\(A=\frac{1009}{2020}:\frac{1}{2}\)
\(A=\frac{1009}{1010}\)
29 tháng 5 2020
a) Ta có
A= 4/2*4+4/4*6+....+4/2018*2020
=> A= 2*(2/2*4+2/4*6+...+2*(2018*2020)
=> A= 2*(1/2-1/4+1/4-1/6+...+1/2018-1/2020)
=> A= 2*(1/2-1/2020)
=> A= 2* 1009/2020
=> A= 1009/1010
b) B= 1/18+1/54+1/108+...+1/990
=> B= 3/3*(1/18+1/54+1/108+..+1/990)
=> B= 1/3*( 3/3*6+3/6*9+...+3/30*33)
=> B= 1/3*(1/3-1/6+1/6-1/9+1/9-1/12+...+1/30-1/33)
=> B= 1/3*( 1/3-1/33)
=> B=1/3*10/33
=> B=10/99
CN
12 tháng 2 2019
\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+....+1\right)=0\)
\(\Leftrightarrow x-2013=0\)(because 1/2012 +1/2011+...+1 luôn lớn hơn 0
\(\Leftrightarrow x=2013\)
Vậy ........
29 tháng 8 2018
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{29}+\frac{1}{30}\)
\(A=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\right)\)
\(A>\left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)\)
\(A>10.\frac{1}{20}+10.\frac{1}{30}\)
\(A>\frac{1}{2}+\frac{1}{3}\)
\(A>\frac{5}{6}\)
Vậy \(A>\frac{5}{6}\)
Chúc bạn học tốt ~
D
29 tháng 8 2018
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{29}+\frac{1}{30}\)
\(A=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}\right)\)
\(A>\left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)\)
\(A>\frac{1}{20}\times10+\frac{1}{30}\times10\)
\(A>\frac{1}{2}+\frac{1}{3}\)
\(A>\frac{5}{6}\)
Vậy \(A>\frac{5}{6}\)