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a)\(\frac{7}{1}.\frac{5}{9}.\frac{11}{35}=\frac{7.5.11}{9.5.7}=\frac{11}{9}\)
b) \(\frac{3}{5}.\frac{4}{23}.\frac{5}{6}=\frac{3.2.2.5}{5.23.3.2}=\frac{2}{23}\)
c) \(\frac{5}{2}.\frac{7}{4}.\frac{2}{7}=\frac{5.7.2}{2.7.4}=\frac{5}{4}\)
d) \(\frac{19}{23}.\frac{12}{17}.\frac{85}{24}=\frac{19.12.5.17}{23.17.12.2}=\frac{19.5}{23.2}=\frac{95}{46}\)
Tính
A. \(\frac{7}{1}x\frac{5}{9}x\frac{11}{35}=\frac{7x5x11}{1x9x35}=\frac{11}{9}\)
B. \(\frac{3}{5}x\frac{4}{23}x\frac{5}{6}=\frac{3x4x5}{5x23x6}=\frac{2}{23}\)
C. \(\frac{5}{2}x\frac{7}{4}x\frac{2}{7}=\frac{5x7x2}{2x4x7}=\frac{35}{28}\)
D. \(\frac{19}{23}x\frac{12}{17}x\frac{85}{24}=\frac{19x12x85}{23x17x24}=\frac{4845}{2346}\)
a,
A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99
=1−(3+5+7+...+99)=1−(3+5+7+...+99)
=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498
b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98
c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99
a) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
=\(1-\dfrac{1}{6}\)=\(\dfrac{5}{6}\)
b) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}\)
=\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)
=\(\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+\dfrac{1.2}{7.9.2}+\dfrac{1.2}{9.11.2}+\dfrac{1.2}{11.13.2}\)
=\(\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\right)\).
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)=\(\dfrac{1}{2}.\dfrac{10}{39}\)=\(\dfrac{5}{39}\).
c) \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
=\(1-\dfrac{1}{8}=\dfrac{7}{8}\).
d) \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4}{2^5}+\dfrac{2^3}{2^5}+\dfrac{2^2}{2^5}+\dfrac{2}{2^5}+\dfrac{1}{2^5}\)
=\(\dfrac{2^4+2^3+2^2+2+1}{2^5}\)=\(\dfrac{2^5-1}{2^5}=\dfrac{31}{32}\).
e) \(\dfrac{1}{7}+\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}=\dfrac{7^{99}+7^{98}+7^{97}+...+7+1}{7^{100}}=\dfrac{\dfrac{7^{100}-1}{6}}{7^{100}}=\dfrac{7^{100}-1}{6.7^{100}}\)
các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha
a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)
-\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)
-6 < \(x\) < -4
vì \(x\) \(\in\) Z nên \(x\) = -5
a\()\) 16/9 +3/5
=107/45
b\()\) 4/13--2/17
=51/221--26/221
=77/221
c\()\) -3/2+4/5
=-15/10+8/10
=-7/10
d\()\) 3/-4-1/4
=-1
e\()\) -1/5.5/7
=-1/7
f\()\) 7/8.64/49
=8/7
g\()\) 3/4.15/24
=15/32
N=7/2(2/1.3+....+2/13.15)
N=7/2.(1/1-1/3+.....+1/13-1/15)
N=7/2.(1-1/15)
N=7/2.(14/15)
N=7.14/2.15
Bạn giải M luôn dc ko?