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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
a)1.2.3.4...9-1.2.3.4...8-1.2.3.4...8.8
=1.2.3.4...8(9-1-8)
=1.2.3.4...8.0
=0
b)(3.4.216)2/11.123.411-169=(3.22.216)2/11.213.222-236=32.24.232/11.235-236=32.226/235.(11-2)
=32.236/235.9=32.236/235.32=2
c)70.(131313/565656+131313/727272+131313/909090
=70.(13/56+13/72+13/90)
=70.39/70=39
d)1/4.9+1/9.14+1/14.19+...+1/64.69
=4/4.9.4+4/9.4.14+4/14.19.4+...+4/64.69.4.
=1/4.(4/4.9+4/9.14+4/14.19+...+4/64.69)
=1/4.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/64-1/69)
=1/4.(1/4-1/69)
=1/4.65/276=65/1104
~~~~~~~~Chúc bạn học giỏi nhé !~~~~~~~~
Áp dụng \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{48.49.50}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{306}{1225}\)
\(1\cdot2\cdot3\cdot...\cdot8\cdot9-1\cdot2\cdot3\cdot...\cdot8-1\cdot2\cdot3\cdot...\cdot8^2\)
=\(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot\left(9-1-8\right)\)(đặt 1*2*3*...*8 ra ngoài)
=\(1\cdot2\cdot3\cdot...\cdot8\cdot0=0\)
\(1.1!+2.2!+3.3!+4.4!+5.5!\\ \)
\(=1.1.1+2.1.2+3.1.2.3+4.1.2.3.4+5.1.2.3.4.5\)
\(=1+2^2.1+3^3.1.2+4^2.1.2.3+5^2.1.2.3.4\)
Ngồi tính :)
\(\Rightarrow A=5\left(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}\right)\)
\(\Rightarrow A=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=5\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=\frac{5x99}{100}=\frac{99}{20}\)
\(A=\frac{5}{1}-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+\frac{5}{3}-\frac{5}{4}+....+\frac{5}{99}-\frac{5}{100}\)
\(A=\frac{5}{1}+\left(-\frac{5}{2}+\frac{5}{2}\right)+\left(-\frac{5}{3}+\frac{5}{3}\right)+\left(-\frac{5}{4}+\frac{5}{4}\right)+...\left(-\frac{5}{99}+\frac{5}{99}\right)+\frac{5}{100}\)
\(A=\frac{5}{1}+0+0+....+0+\frac{5}{100}\)
\(A=\frac{500}{100}+\frac{5}{100}=\frac{205}{100}=\frac{101}{20}\)
Đúng 100%
Đúng 100%
Đúng 100%
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2014.2015.2016}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2014.2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2014.2015}-\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2015.2016}\right)\)
Bài giải chi tiết:
Ta có:
S = 1 x 2 + 2 x 3 + 3 x 4 + ...+ 38 x 39 + 39 x 40
S x 3 = 1 x 2x 3 + 2 x 3x 3 + 3 x 4x 3 +… + 38 x 39 x 3 + 39 x 40 x 3
S x 3 = 1 x 2 x 3 + 2 x 3 x (4 - 1) + 3 x 4 x (5-2) + ... + 38 x 39 x (40 - 37) + 39 x40 x(41 - 38)
S x 3 = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + ... + 38 x 39 x 40 - 37x 38 x 39 + 39 x 40 x 41 - 38 x 39 x 40.
S x 3 = 39 x 40 x 41
S = 39 x 40 x 41 : 3= 21320
Bỏ số 40 nhỏ ở cuối đi