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\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=308-3\)
\(x=305\)
Vậy \(x=305\)
Tham khảo nhé~
\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
<=>\(\frac{1}{x+3}=\frac{1}{308}\)
<=> x+3=308
<=> x=305
\(\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(3\left(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{3\cdot101}{1540}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{308}{1540}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)
\(x+3=308\)
\(x=308-3=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{308}{1540}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{5}{1540}=\frac{1}{308}\)
=> x + 3 = 308
=> x = 308 - 3
=> x = 305
Vậy x = 305
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{5}{1540}=\frac{1}{308}\)
=> x + 3 = 308
=> x = 308 - 3
=> x = 305
Vậy x = 305
còn chi tiết đây
a)1/5.8+1/8.11+1/11.14+...+1/x.(x+3)=101/1540
1/(5.8)+1/(8.11)+1/(11.14)+...1/x.(… =101/1540
3/(5.8)+3/(8.11)+...+3/x(x+3)=3.(10…
1/5-1/8+1/8-1/11+...+1/x-1/(x+3)=30…
1/5-1/(x+3)=303/1540
1/(x+3)=1/5-303/1540=1/308
=>x=305
lời giải nè : ấn vô dòng đen đen ở dưới ấy nhé
Tìm x, biết:a) 1/5.8 + 1/8.11 + 1/11.14 + ... + 1/x.(x+3)= 101/1540b) 1+ 1/3 + 1/6 + 1/10 +...+ 1/x.(x+1):2 = $1\frac{1991}{1993}$119911993
\(3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(3.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{4620}\)
\(\frac{1}{x+3}=...\) (tự làm tiếp)
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+1\right)}=\frac{101}{1540}\)
\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{101}{1540}\)
\(\frac{1}{5}-\frac{1}{x+1}=\frac{101}{1540}:\frac{1}{3}=\frac{303}{1540}\)
\(\frac{1}{x+1}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=> x + 1 = 308
=> x = 308 - 1
=> x = 307
1) \(15^{40}=15^{4.10}=\left(15^4\right)^{10}=50625^{10}\)
\(18^{30}=18^{3.10}=\left(18^3\right)^{10}=5832^{10}\)
Vì \(50625^{10}>5832^{10}\) Nên \(15^{40}>18^{30}\)
2)\(180^{300}=180^{3.100}=\left(18^3\right)^{100}=5832^{100}\)
Vì \(2015^{100}