Tìm x biết : \(x\times\left(\frac{2015}{8\times9}+\frac{1925}{9\times10}+\frac{1795}{10\times11}+\frac{1629}{11\times12}+6\right)=\frac{1}{24}\)
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1/5x8 + 1/8x11 + 1/11x14 + ... + 1/xx(x+3) = 101/1540
1/3 x (3/5x8 + 3/8x11 + 3/11x14 + ... + 3/xx(x+3) = 101/1540
1/3 x (1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + ... + 1/x - 1/x+3) = 101/1540
1/3 x (1/5 - 1/x+3) = 101/1540
1/5 - 1/x+3 = 101/1540 : 1/3
1/5 - 1/x+3 = 303/1540
1/x+3 = 1/5 - 303/1540
1/x+3 = 1/308
=> x+3=308
=> x=308-3=305
vậy x=305
1/5x8 + 1/8x11 + 1/11x14 + ... + 1/xx(x+3) = 101/1540
1/3 x (3/5x8 + 3/8x11 + 3/11x14 + ... + 3/xx(x+3) = 101/1540
1/3 x (1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + ... + 1/x - 1/x+3) = 101/1540
1/3 x (1/5 - 1/x+3) = 101/1540
1/5 - 1/x+3 = 101/1540 : 1/3
1/5 - 1/x+3 = 303/1540
1/x+3 = 1/5 - 303/1540
1/x+3 = 1/308
=> x+3=308
=> x=308-3=305
vậy x=305
\(\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\frac{10}{11}\times y=\frac{2}{3}\)
\(\frac{5}{11}\times y=\frac{2}{3}\) => \(y=\frac{2}{3}:\frac{5}{11}=\frac{2}{3}\times\frac{11}{5}=\frac{22}{15}\)
\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)
\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)
\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)
\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)
\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)
Đến đây là tính dễ rồi :v
\((-3,2)\cdot\frac{-15}{64}+\left[0,8-2\frac{4}{15}\right]:1\frac{23}{24}\)
\(=\frac{-32}{10}\cdot\frac{-15}{64}+\left[\frac{8}{10}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-32\cdot(-15)}{10\cdot64}+\left[\frac{4}{5}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-1\cdot(-3)}{2\cdot2}+\frac{4\cdot3-34}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-22}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-517}{180}=\frac{-191}{90}\)
Bài 2 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
\(x\times\left(\frac{2015}{8\times9}+\frac{1925}{9\times10}+\frac{1795}{10\times11}+\frac{1629}{11\times12}+6\right)=\frac{1}{24}\)
=> \(x\times\left(\frac{2015}{72}+\frac{1925}{90}+\frac{1795}{110}+\frac{1629}{132}+6\right)=\frac{1}{24}\)
=> \(x\times84\frac{3}{88}=\frac{1}{24}\Rightarrow x=\frac{1}{24}:\frac{7395}{88}=\frac{1}{24}\times\frac{88}{7395}=\frac{88}{177480}=\frac{11}{22185}\)