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B=\(\dfrac{1+15^4+15^8+...+15^{96}+15^{100}}{\left(1+15^4+15^8+..+15^{96}+15^{100}\right)+\left(15^2+15^6+...+15^{98}+15^{102}\right)}\)
=\(\dfrac{1+15^4+15^8+...+15^{96}+15^{100}}{\left(1+15^4+15^8+...+15^{96}+15^{100}\right)+15^2.\left(1+15^{14}+15^8+...+15^{96}+15^{100}\right)}\)
\(\dfrac{\left(1+15^4+15^8+...+15^{96}+15^{100}\right)}{\left(1+15^4+15^8+...+15^{96}+15^{100}\right)\left(1+15^2\right)}\)
=\(\dfrac{1}{1+15^2}=\dfrac{1}{226}\)
Bai 1
đặt A = 1 + 15^4 + 15^8 + .... + 15^100
=> 15^4A = 15^4 + 15^8 + 15^12 + .... + 15^104
ta có
15^4A = 15^4 + 15^8 + 15^12 + .... + 15^100 + 15^104
-
A = 15^4 + 15^8 + 15^12 + .... + 15^100 + 1
50624A = 15^104 - 1
=> A = (15^104-1)/50624
bài 2 làm tương tự cũng đặt A và nhân A với 15^4 (bạn thông cảm mình không có nhiều thời gian)
\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)
\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)
\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)
\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)
\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)
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 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
A/ \(\left(15-6\frac{13}{18}\right):11\frac{1}{27}-2\frac{1}{8}:1\frac{11}{40}\)
\(=\left(15-\frac{121}{18}\right):\frac{298}{27}-\frac{17}{8}:\frac{51}{40}\)
\(=\left(\frac{270}{18}-\frac{121}{18}\right):\frac{298}{27}-\frac{17}{8}:\frac{51}{40}\)
\(=\frac{149}{18}:\frac{298}{27}-\frac{17}{8}:\frac{51}{40}\)
\(=\frac{3}{4}-\frac{5}{3}\)
\(=\frac{9}{12}-\frac{20}{12}\)\(=-\frac{11}{12}\)
B/ \(\left(-3,2\right)\cdot-\frac{15}{64}+\left(0,8-2\frac{4}{15}\right):3\frac{2}{3}\)
\(=\left(-3,2\right)\cdot-\frac{15}{64}+\left(0,8-\frac{34}{15}\right):\frac{11}{3}\)
\(=-\frac{3,2}{1}\cdot-\frac{15}{64}+\left(0,8-\frac{34}{15}\right):\frac{11}{3}\)
\(=\frac{48}{64}+\left(0,8-\frac{34}{15}\right):\frac{11}{3}\)
\(=\frac{3}{4}+\left(\frac{12}{15}-\frac{34}{15}\right):\frac{11}{3}\)
\(=\frac{3}{4}+\left(-\frac{22}{15}\right):\frac{11}{3}\)
\(=\frac{3}{4}+\left(-\frac{2}{5}\right)\)
\(=\frac{15}{20}+\left(-\frac{8}{20}\right)\)
\(=\frac{7}{20}\)