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\(\frac{x+1}{3}=\frac{9}{2}\)
\(\left(x+1\right).2=9.3\)
\(\left(x+1\right).2=27\)
\(x+1=27:2\)
\(x+1=13,5\)
\(x=13,5-1=12,5\)
vậy x = 12.5
\(\frac{x+1}{3}=\frac{9}{2}\)
\(\Leftrightarrow2\left(x+1\right)=3\times9\)
\(\Leftrightarrow2\left(x+1\right)=27\)
\(\Leftrightarrow x+1=\frac{27}{2}\)
\(\Leftrightarrow x=\frac{25}{2}\)
\(b,\left(2\chi-7\right)^{4-1}=4^{2\times5}\)\(a,3\times2^{\chi-7}=17\)
a) \(3.2^x-7=17\)
\(3\cdot2^x=24\)
\(2^x=8=2^3\)
=> x = 3
b) \(\left(2x-7\right)^4-1=4^2\cdot5\)
\(\left(2x-7\right)^4-1=80\)
\(\left(2x-7\right)^4=81=\left(\pm3\right)^4\)
+) 2x - 7 = 3
2x = 10
x = 5
+) 2x - 7 = -3
2x = 4
x = 2
Vậy,...........
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow A=\frac{5}{11}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\Rightarrow B=\frac{1009}{2019}\)
\(\frac{2}{7}C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\Rightarrow C=\frac{2018}{2019}:\frac{2}{7}=\frac{7063}{2019}\)
\(\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{100.103}\)
\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{100}-\frac{1}{103}\)
\(=\frac{1}{7}-\frac{1}{103}\)
\(=\frac{96}{721}\)
\(\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(=\frac{2}{3}.\frac{96}{721}\)
\(=\frac{64}{721}\)
\(A=\)\(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(B=2\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2.3\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(3B=2.\frac{96}{721}\)
\(3B=\frac{192}{721}\)
\(\Rightarrow B=\frac{192}{721}:3\)
\(B=\frac{64}{721}\)
\(A=\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
Vậy \(A=\frac{96}{721}\)
\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(B=\frac{2}{3}.\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)
\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}.\frac{96}{721}\)
\(B=\frac{64}{721}\)
Vậy \(B=\frac{64}{721}\)
_Chúc bạn học tốt_
A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)
A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)
A=3.(1-1/400)
A=3.399/400
A=1197/400
A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)
A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)
A=3.(1-1/400)
A=3.399/400
A=1197/400
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