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A = -4/5x(1/2+1/3+1/4)= -4/5x1 = -4/5
B = 6/19 x ( 3/4+4/3+-1/2)= 6/19x 19 = 6
C = 2002/2003x(3/4+5/6-19/12)=2003/2002x0=0
a) \(\left(\frac{11}{4}.\frac{-5}{9}-\frac{4}{9}.\frac{11}{4}\right).\frac{8}{33}\)
=\(\frac{11}{4}\left(-\frac{5}{9}-\frac{4}{9}\right).\frac{8}{33}\)
=\(\frac{11}{4}\cdot-1\cdot\frac{8}{33}\)
=\(-\frac{11}{4}\cdot\frac{8}{33}\)
=\(-\frac{2}{3}\)
b)\(-\frac{1}{4}\cdot\frac{152}{11}+\frac{68}{4}\cdot-\frac{1}{11}\)
=\(\frac{-1.152}{4.11}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1.152}{11.4}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1}{11}\cdot\frac{152}{4}+\frac{68}{4}\cdot\frac{-1}{11}\)
=\(\frac{-1}{11}\cdot\left(\frac{152}{4}+\frac{68}{4}\right)\)
=\(\frac{-1}{11}\cdot55=-5\)
c)\(\frac{-2}{3}\cdot\frac{4}{5}+\frac{2}{3}\cdot\frac{3}{5}\)
=\(-1\cdot\frac{2}{3}\left(\frac{4}{5}+\frac{3}{5}\right)\)
=\(-1\cdot\frac{2}{3}\cdot\frac{7}{5}\)
=\(-\frac{2}{3}\cdot\frac{7}{5}\)
=\(\frac{-14}{15}\)
d) chưa nghĩ ra nhé
e) bạn chép sai đề bài rồi
mk mới kiểm tra 45 phút nên biết
đề bài nè
\(\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot...\cdot\frac{899}{30^2}\)
=\(\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}\cdot\frac{3.5}{4^2}\cdot...\cdot\frac{29.31}{30^2}\)
=\(\frac{1.3.2.4.3.5...29.31}{2.2.3^2.4^2...30.30}\)
=\(\frac{1.2.3^2.4^2.5^2....29^2.30.31}{2.2.3^2.4^2.5^2....29^2.30.30}\)
=\(\frac{1.31}{2.30}\)
=\(\frac{31}{60}\)
a)trong ngoac bn dat thau so chung la 11/4 rui tinh binh thuong b)bn tu lam nhe c)dat thua so chung d)tinh trong ngoac ra rui nhan vs e) mk bo tay
M=(1.3.5.7.....99)/(2.4.6.8.....100)
số số hạng của tử = (99-1)/2 +1 = 50 -> 1.3.5.7....99= (99+1)*50/2 =2500
số số hạng của mẫu = (100-2)/2+1 =50 -> 2.4.6.8....100= (100+2)*50/2 =2550
--> M= 2500/2550 =50/51
Làm tương tự với N ta có kq N=51/52 ->M/N= 2600/2601 -> M<N
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)
= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)
b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)
\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)
\(=66+44+33+22+12=177\)
c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)
= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)
= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)
= còn lại làm nốt nha! bận ròy
\(\frac{1}{n\cdot (n+1)\cdot (n+2)}=\frac{1}{2}\cdot \left[\frac{1}{n\cdot (n+1)}-\frac{1}{(n+1)\cdot (n+2)}\right]\)
\(2S=\frac{2}{1\cdot 2\cdot 3}+\frac{2}{2\cdot 3\cdot 4}+...+\frac{2}{98\cdot 99\cdot 100}\)
\(2S=\left(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}\right)+\left(\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}\right)+...+\left(\frac{1}{98\cdot 99}-\frac{1}{99\cdot 100}\right)\)
\(2S=\frac{1}{1\cdot 2}-\frac{1}{99\cdot 100}=\frac{1}{2}-\frac{1}{9900}\)
\(2S=\frac{4950}{9900}-\frac{1}{9900}=\frac{4949}{9900}\)
\(S=\frac{4949}{9900}:2=\frac{4949}{19800}\)
S = 1/(1.2.3) + 1/(2.3.4) + ... + 1/(98.99.100)
Ta có 1/[n(n + 1)(n + 2)] = 1/2.[1/(n(n + 1)) - 1/((n + 1)(n + 2))]
S = 1/2.(1/(1.2) - 1/(99.100))
S = 1/2.(1/2 - 1/9900)
S = 1/2.4949/9900
S = 4949/19800
Vậy S = 4949/19800, vì tổng rút gọn theo dạng khử liên tiếp.
`S = 1/(1*2*3) + 1/(2*3*4) + ..... + 1/(98*99*100)`
`=>2S = 2/(1*2*3) + 2/(2*3*4) + ... + 2/(98*99*100)`
`=>2S = 1/(1*2) - 1/(2*3) + 1/(2*3) - 1/(3*4) + ... + 1/(98*99) - 1/(99*100)`
`=> 2S = 1/2 - 1/(99*100)`
`=> 2S = 1/2 - 1/9900`
`=> 2S = 4949/9900`
`=>S = 4949/19800`
Vậy `S = 4949/19800`
\(\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\)Các bước giải chi tiết
\(S=\frac{1}{2}\left(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}\right)+\frac{1}{2}\left(\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}\right)+\dots +\frac{1}{2}\left(\frac{1}{98\cdot 99}-\frac{1}{99\cdot 100}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}+\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}+\dots +\frac{1}{98\cdot 99}-\frac{1}{99\cdot 100}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{1\cdot 2}-\frac{1}{99\cdot 100}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(S=\frac{1}{2}\left(\frac{4950}{9900}-\frac{1}{9900}\right)\)
\(S=\frac{1}{2}\cdot \frac{4949}{9900}=\frac{\mathbf{4949}}{\mathbf{19800}}\)