E=1-2+3-4+5-6+...+71-72
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TN
5 tháng 7 2016
a)×+1/ 53 + ×+2 /52 + ×+3/ 51+3 = 0
\(\Rightarrow\frac{x+1}{53}+1+\frac{x+2}{52}+1+\frac{x+3}{51}+1+\frac{3\left(x+54\right)}{\left(x+54\right)}=0\)
\(\Rightarrow\frac{x+54}{53}+\frac{x+54}{52}+\frac{x+54}{51}+\frac{x+54}{\frac{1}{3}\left(x+54\right)}=0\)
\(\Rightarrow\left(x+54\right)\left(\frac{1}{53}+\frac{1}{52}+\frac{1}{51}+\frac{1}{\frac{1}{3}\left(x+54\right)}\right)=0\)
\(\Rightarrow x+54=0\).Do \(\frac{1}{53}+\frac{1}{52}+\frac{1}{51}+\frac{1}{\frac{1}{3}\left(x+54\right)}\ne0\)
=>x=-54
b)×-2/ 72 + ×-3/ 71 + ×-4/ 70 -3 = 0
\(\Rightarrow\frac{x-2}{72}-1+\frac{x-3}{71}-1+\frac{x-4}{70}-1-\frac{3\left(x-74\right)}{x-74}=0\)
\(\Rightarrow\frac{x-74}{72}+\frac{x-74}{71}+\frac{x-74}{70}-\frac{x-74}{\frac{1}{3}\left(x-74\right)}=0\)
\(\Rightarrow\left(x-74\right)\left(\frac{1}{72}+\frac{1}{71}+\frac{1}{70}-\frac{1}{\frac{1}{3}\left(x-74\right)}\right)=0\)
\(\Rightarrow x-74=0\).Do \(\frac{1}{72}+\frac{1}{71}+\frac{1}{70}-\frac{1}{\frac{1}{3}\left(x-74\right)}\ne0\)
=>x=74
c)×+5/ 81 + ×+4/ 41 + ×-7/ 31 + 6 = 0
\(\Rightarrow\frac{x+5}{81}+1+\frac{x+4}{41}+2+\frac{x-7}{31}+3+\frac{6\left(x+86\right)}{x+86}=0\)
\(\Rightarrow\frac{x+86}{81}+\frac{x+86}{41}+\frac{x+86}{31}+\frac{x+86}{\frac{1}{6}\left(x+86\right)}=0\)
\(\Rightarrow\left(x+86\right)\left(\frac{1}{81}+\frac{1}{41}+\frac{1}{31}+\frac{1}{\frac{1}{6}\left(x+86\right)}\right)=0\)
\(\Rightarrow x+86=0\).Do \(\frac{1}{81}+\frac{1}{41}+\frac{1}{31}+\frac{1}{\frac{1}{6}\left(x+86\right)}\ne0\)
=>x=-86
d)tương tự nhé
ND
1
20 tháng 8 2021
Ta có: \(\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\)
\(=8-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=8-\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=8-1+\dfrac{1}{10}\)
\(=\dfrac{71}{10}\)
ND
Nguyễn Đức Trí
VIP
8 tháng 9 2023
\(G=1+2012+2012^2+2012^3+2012^4+...+2012^{71}+2012^{72}\)
\(\Rightarrow G=\dfrac{2012^{72+1}-1}{2012-1}\)
\(\Rightarrow G=\dfrac{2012^{73}-1}{2011}< H=2012^{73}-1\)
VN
0
T
25 tháng 8 2023
\(A=\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}+\dfrac{71}{72}+\dfrac{89}{90}\) (sửa đề)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+\left(1-\dfrac{1}{30}\right)+\left(1-\dfrac{1}{42}\right)+\left(1-\dfrac{1}{56}\right)+\left(1-\dfrac{1}{72}\right)+\left(1-\dfrac{1}{90}\right)\)
\(=\left(1+1+1...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=8-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\) ( có 8 số hạng 1)
\(=8-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=8-\left(1-\dfrac{1}{10}\right)\)
\(=8-\dfrac{9}{10}\)
\(=\dfrac{80}{10}-\dfrac{9}{10}=\dfrac{71}{10}\)
25 tháng 8 2023
A=1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90
=1−1/2+1−1/6+1−1/12+1−1/20+1−1/30+1−1/42+1−1/56+1−1/72+1−1/90
=9−(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)
=9−(1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10)
=9-(1-1/2+1/2-1/3+.....+1/9-1/10)
=9−(1−1/10)
=9−1+1/10=8+1/10=81/10
DA
16 tháng 8 2017
a)\(A=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}\)
\(\frac{1}{2}xA=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(\frac{1}{4}xA=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}\)
\(\frac{1}{4}xA-\frac{1}{2}xA=\frac{1}{3}-\frac{1}{384}\)
\(\frac{1}{4}xA=\frac{127}{384}\)
\(A=\frac{127}{384}:\frac{1}{4}\)
\(A=\frac{127}{96}\)
KT
15 tháng 10 2018
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
Tính
\(E=1-2+3-4+5-6+...+71-72\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(71-72\right)\) (có 36 cặp)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=\left(-1\right).36=-36\)
Vậy \(E=-36\).
Ta có: E=1-2+3-4+5-6+...+71-72
=> E=(1-2)+(3-4)+(5-6)+...+(71-72)
=> E= (-1)+(-1)+(-1)+...+(-1)
Dãy trên có số sô hạng là: (72-1):1+1=72 (số hạng)
Có số cặp là: 72:2=36(cặp)
=> E=(-1) x 36=-36