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XO
27 tháng 9 2020
Ta có : \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
=> \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)
=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1989}{1991}\)
=> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1989}{3982}\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1989}{3982}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1989}{3982}\)
=> \(\frac{1}{x+1}=\frac{1}{1991}\)
=> x + 1 = 1991
=> x = 1990
Vậy x = 1990
27 tháng 9 2020
\(2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{1990}{1991}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1990}{1991}\)
\(1-\frac{1}{x+1}=\frac{1990}{1991}\)
\(\frac{1}{x+1}=1-\frac{1990}{1991}\)
\(\frac{1}{x+1}=\frac{1}{1991}\)
\(x+1=1991\)
\(x=1990\)
LM
7 tháng 9 2016
a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)
b.
\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)
\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)
\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)
\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)
22 tháng 9 2016
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(=3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x.3}=\frac{303}{1540}\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{308}\)
\(x+3=308\)
\(\Rightarrow x=305\)
17 tháng 8 2018
a,(=)\(3^{x+1}.\left(3+4\right)=7.3^6\)
(=)\(3^{x+1}=3^6\)
=>x+1=6(=)x=5
b
T
5 tháng 6 2019
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
T
5 tháng 6 2019
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.13}+....+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\Rightarrow x=305\)
câu 2 ko cần làm đâu
1 ) \(A=\frac{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}}{\frac{1}{99.1}+\frac{1}{97.3}+...+\frac{1}{3.97}+\frac{1}{1.99}}\)
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+....+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{1.99}+\frac{100}{3.97}+....+\frac{100}{49.51}\)
\(=100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)\)
\(\Rightarrow A=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+....+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+....+\frac{1}{49.51}\right)}=\frac{100}{2}=50\)
Bài 3 thiếu vế phải nha !!!!
\(\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{x\left(x+3\right)}\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Bài 4 đề ko dõ dàng !!!!
bài 5 ) \(1+5+9+13+...+x=5050\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(\frac{x-1}{4}+1\right)}{2}=5050\)
\(\Leftrightarrow\frac{\left(x+1\right).\frac{x+3}{4}}{2}=5050\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{x+3}{4}\right)=10100\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=40400\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=200.202=\left(199+1\right)\left(199+3\right)\)
\(\Rightarrow x=199\)
A = 2009 + 20092 + 20093 + ........ + 200910
= (2009 + 20092) + (20093 + 20094) + ........ + (20099 + 200910)
= 2009(1 + 2009) + 20093(1 + 2009) + ........ + 20099(1 + 2009)
= 2009.2010 + 20093.2010 + ....... + 20099.2010
= 2010(2009 + 20093 + ....... + 20099 ) chia hết cho 2010
=> A = 2009 + 20092 + 20093 + ........ + 200910 chia hết cho 2010
câu 3 =101/1540
thank you very much
Câu 4 ) \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x\left(x+1\right)}:2=1\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{x\left(x+1\right)}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow x+1=2\Rightarrow x=1\)
Với vế phải là 1991/1943 thì làm tương tự nhé , mình mỏi tay quá
sai câu 4 ý là 1\(\frac{1991}{1540}\)
ý là 1\(\frac{1991}{1540}\)