\(\frac{72+36\times2+24\times3+18\times4+12\times6+168}{2+2+4+6+...+512+1024}\)
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Ta có :
\(S=\frac{5\times2^{30}\times6^2\times3^{15}-2^3\times8^9\times3^{17}\times21}{21\times2^{29}\times3^{16}\times4-2^{29}\times\left(3^4\right)^5}\)
\(S=\frac{5\times2^{30}\times2^2\times3^2\times3^{15}-2^3\times2^{27}\times3^{17}\times3\times7}{3\times7\times2^{29}\times3^{16}\times2^2-2^{29}\times3^{20}}\)
\(S=\frac{5\times2^{32}\times3^{17}-2^{30}\times3^{18}\times7}{7\times2^{31}\times3^{17}-2^{29}\times3^{20}}\)
\(S=\frac{2^{30}\times3^{17}\times\left(5\times2^2-3\times7\right)}{2^{29}\times3^{17}\times\left(2^2\times7-3^3\right)}\)
\(S=\frac{2^{30}\times3^{17}\times\left(-1\right)}{2^{29}\times3^{17}\times1}\)
\(\Rightarrow S=-2\)
Ko viết đề :)
\(S=\frac{5\cdot2^{30}\cdot2^2\cdot3^2\cdot3^{15}-2^3\cdot2^{27}\cdot3^{17}\cdot3\cdot7}{3\cdot7\cdot2^{29}\cdot3^{16}\cdot2^2-2^{29}\cdot3^{20}}\)
\(=\frac{5\cdot2^{32}\cdot3^{17}-2^{30}\cdot3^{18}\cdot7}{3^{17}\cdot7\cdot2^{31}-2^{29}\cdot3^{20}}\)
\(=\frac{2^{30}\cdot3^{17}\left(5\cdot2^2-3\cdot7\right)}{2^{29}\cdot3^{17}\left(7\cdot2^2-3^3\right)}\)
\(=\frac{2\left(20-21\right)}{28-27}\)
\(=\frac{40-42}{1}=-\frac{2}{1}=-2\)
Vậy S= -2
dau . la dau x
a/ 1.3.2.4.3.5.4.6.5.7/2.2.3.3.4.4.5.5.6.6=1.7/2.6=7/12
b/ ab.aba=abab
aba=abab:ab
aba=101
=>a=1 b=0
aabb : ab = 99 hay ab x 99 = aabb hay ab x100 – ab = aabb
Ta có phép tính
__ ab00
___ab___
aabb
b=0 hoặc b=5
Nếu b=0 thì a000 – a0 = aa00 (sai)
Nếu b=5 thì
__ a500
__a5___
aa55
a=4
c) thay a=7/6 b=6/5 thi 3 x a + 4 : b - 5/12=3.7/6+4.6/5-5/12=7/2+24/5-5/12=210/60+288/60-25/60=473/60
**** nha
\(\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{\left(2.3.4.5.6\right).\left(3.4.5.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{7}{12}\)
Áp dụng \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{306}{1225}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)