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\(M=\frac{1.2.3.4.5...98.99}{10}\)
\(M=1.2.3.4.5.6.7.8.9.11.12...98.99\)
\(C=\frac{1.5.6+2.10.12+24.8.10}{1.3.5+2.6.10+8.6.20}\)
\(C=\frac{1.5.6.\left(1^3+2^3+8^2\right)}{1.3.5.\left(1^3+2^3+8^2\right)}=\frac{6}{3}=2\)
\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}\cdot\frac{17}{4}-28\cdot\frac{4}{3}\right):\frac{7}{4}\)
\(=\frac{59}{15}-\frac{29}{4}:\frac{7}{4}=\)\(\frac{59}{15}-\frac{29}{7}=\frac{-22}{105}\)
B = \(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{17}{4}-2x\frac{4}{3}\right):\frac{7}{4}\)
= \(\frac{59}{10}x\frac{2}{3}-\left(\frac{119}{12}-\frac{8}{3}\right)x\frac{4}{7}\)
= \(\frac{59}{15}-\frac{29}{4}x\frac{4}{7}=\frac{59}{15}-\frac{29}{7}\)
= \(\frac{-22}{105}\)
C = \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)
= \(1-\frac{1}{7}=\frac{6}{7}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
= 1 / 1 - 1 / 2 + 1 / 2 - 1 / 3 + 1 / 3 - 1 / 4 + 1 / 4 - 1 / 5 + 1 / 5 - 1 / 6
Ta gạch các ps trùng.
Còn lại :
1 / 1 - 1 / 6 = 6 / 5
a) \(T=\frac{9^{14}\times25^6\times8^7}{18^{12}\times625^3\times24^3}\)
\(=\frac{\left(3^2\right)^{14}\times25^6\times\left(2^3\right)^7}{\left(2\times3^2\right)^{12}\times\left(25^2\right)^3\times\left(3\times2^3\right)^3}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{2^{12}\times3^{24}\times25^6\times3^3\times2^9}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{\left(2^{12}\times2^9\right)\times\left(3^{24}\times3^3\right)\times25^6}\)
\(=\frac{3^{28}\times25^6\times2^{21}}{2^{21}\times3^{27}\times25^6}=3\)
b) \(A=\frac{5\times4^{15}\times9^9-4\times3^{20}\times8^9}{5\times2^9\times6^{19}-7\times2^{29}\times27^6}\)
\(=\frac{5\times\left(2^2\right)^{15}\times\left(3^2\right)^9-2^2\times3^{20}\times\left(2^3\right)^9}{5\times2^9\times\left(2\times3\right)^{19}-7\times2^{29}\times\left(3^3\right)^6}\)
\(=\frac{5\times2^{30}\times3^{18}-2^2\times3^{20}\times2^{27}}{5\times2^9\times2^{19}\times3^{19}-7\times2^{29}\times3^{18}}\)
\(=\frac{5\times2^{30}\times3^{18}-2^{29}\times3^{20}}{5\times2^{28}\times3^{19}-7\times2^{29}\times3^{18}}\)
\(=\frac{2^{29}\times3^{18}\times\left(5\times2-3^2\right)}{2^{28}\times3^{18}\times\left(5\times3-7\times2\right)}\)
\(=\frac{2\times\left(10-9\right)}{15-14}=\frac{2\times1}{1}=2\)
\(A=\frac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\)
\(=\frac{2\cdot1\cdot3\cdot3\cdot4\cdot2+3\cdot1\cdot4\cdot3\cdot5\cdot2+...+98\cdot1+99\cdot3+100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=\frac{1\cdot3\cdot2\cdot\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=1\cdot3\cdot2\)
\(=6\)
\(A^2=6^2=36\)
\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)
\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)
\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)
\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)
\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)
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 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{3.5.7}+...+\frac{1}{45.47.49}\)
\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{45.47.49}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{45.47}-\frac{1}{47.49}\)
\(=\frac{1}{3}-\frac{1}{47.49}\)
\(\Rightarrow A=\frac{\frac{1}{3}-\frac{1}{47.49}}{4}=\frac{575}{6909}\)
1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6+1/7-1/7+1/8-1/8+1/9+1/9-1/10
=1/2-1/10
=5/10-1/10
=4/10=2/5
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}-\frac{1}{10}\)
\(\frac{2}{5}\)