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\(b,\left(2\chi-7\right)^{4-1}=4^{2\times5}\)\(a,3\times2^{\chi-7}=17\)
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,...........
\(A=\)\(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(B=2\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2.3\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(3B=2.\frac{96}{721}\)
\(3B=\frac{192}{721}\)
\(\Rightarrow B=\frac{192}{721}:3\)
\(B=\frac{64}{721}\)
\(A=\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
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_
\(\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{100.103}\)
\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{100}-\frac{1}{103}\)
\(=\frac{1}{7}-\frac{1}{103}\)
\(=\frac{96}{721}\)
\(\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(=\frac{2}{3}.\frac{96}{721}\)
\(=\frac{64}{721}\)
\(\left(x-1\right)^x=10^{11}\)
\(\Leftrightarrow\left(x-1\right)^x=\left(11-1\right)^{11}\)
\(\Rightarrow x=11\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow A=\frac{5}{11}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\Rightarrow B=\frac{1009}{2019}\)
\(\frac{2}{7}C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\Rightarrow C=\frac{2018}{2019}:\frac{2}{7}=\frac{7063}{2019}\)
9.(2x-20)+2x=-580
=>18x-180+2x=-580
=>18x+2x=(-580)+180
=>20x=-400
=>x=(-400):20
=>x=-20
9.(2x-20)+2x = -580
18x-180+2x=-580
18x+2x-180=-580
20x-180=-580
20x=-580+180
20x=-400
x=-400:20
x=-20
Vậy x=-20
9.(2x-20)+2x=-580
=>18x-180+2x=-580
=>18x+2x=(-580)+180
=>20x=-400
=>x=(-400):20
=>x=-20
9.(2x-20)+2x = -580
18x-180+2x=-580
18x+2x-180=-580
20x-180=-580
20x=-580+180
20x=-400
x=-400:20
x=-20
Vậy x=-20
