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E=\(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\\ E=\dfrac{1}{90}-\left(\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}+\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{2}\right)\\ E=\dfrac{1}{90}-\left(\dfrac{1}{9.8}+\dfrac{1}{8.7}+\dfrac{1}{7.6}+\dfrac{1}{6.5}+\dfrac{1}{5.4}+\dfrac{1}{4.3}+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\\ E=\dfrac{1}{90}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\\ E=\dfrac{1}{90}-\left(1-\dfrac{1}{9}\right)\\ E=\dfrac{1}{90}-\dfrac{8}{9}\\ E=\dfrac{1}{90}-\dfrac{80}{90}\\ E=-\dfrac{79}{90}\)Vậy:\(E=-\dfrac{79}{90}\)
E=\(\dfrac{1}{10.9}-\dfrac{1}{9.8}-\dfrac{1}{8.7}-\dfrac{1}{7.6}-\dfrac{1}{6.5}-\dfrac{1}{5.4}-\dfrac{1}{4.3}-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
E=\(\dfrac{1}{10}-\dfrac{1}{1}\)
E=\(\dfrac{-9}{10}\)
1) \(x+\dfrac{30}{100}x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)
\(\Leftrightarrow100x+30x=-131\)
\(\Leftrightarrow130x=-131\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
Vậy \(x=-\dfrac{131}{130}\)
b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)
\(\Leftrightarrow-72+32x=77\)
\(\Leftrightarrow32x=77+72\)
\(\Leftrightarrow32x=149\)
\(\Leftrightarrow x=\dfrac{149}{32}\)
Vậy \(x=\dfrac{149}{32}\)
Ta có: \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{10}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{19}+\dfrac{1}{20}-1-\dfrac{1}{2}-...-\dfrac{1}{10}\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
Vậy \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
B=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
B=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)
B=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
B= 1-\(\dfrac{1}{8}\)
B= \(\dfrac{7}{8}\)
\(A=\dfrac{5}{9}-\dfrac{5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\ =\dfrac{5}{9}+\dfrac{-5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\= \left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{-5}{8}+\dfrac{-3}{8}\right)\\ =1+1+\left(-1\right)\\ =2+\left(-1\right)\\ =1\)
=>(y-1/2):(1-1/2+1/2-1/3+...+1/9-1/10)=1/6
=>(y-1/2):9/10=1/6
=>(y-1/2)=1/6*9/10=9/60=3/20
=>y=3/20+10/20=13/20
A=1+2+3+4+5+...+99+100
A=\(\dfrac{100.\left(100+1\right)}{2}\)=5050
Vậy A=5050
B=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
B=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
B=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B=\(1-\dfrac{1}{100}\)=\(\dfrac{99}{100}\)
Vậy B=\(\dfrac{99}{100}\)
Chuẩn ồi còn j
Đăng ít thôi.
d) \(D=\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+\dfrac{1}{5.6.7}+\dfrac{1}{6.7.8}+\dfrac{1}{7.8.9}+\dfrac{1}{8.9.10}\)
\(\Rightarrow2D=\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}+\dfrac{2}{7.8.9}+\dfrac{2}{8.9.10}\)
\(\Rightarrow2D=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{4.5}-\dfrac{1}{5.6}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(\Rightarrow2D=\dfrac{1}{2.3}-\dfrac{1}{9.10}\)
\(\Rightarrow2D=\dfrac{22}{45}\)
\(\Rightarrow D=\dfrac{11}{45}\)
a, Ta có :
\(\dfrac{1}{6}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{5}\)
.................
\(\dfrac{1}{9}< \dfrac{1}{5}\)
\(\dfrac{1}{10}=\dfrac{1}{10}\)
\(\dfrac{1}{11}< \dfrac{1}{10}\)
..................
\(\dfrac{1}{17}< \dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+......+\dfrac{1}{17}< \dfrac{1}{5}+\dfrac{1}{5}+....+\dfrac{1}{5}\)
\(\Leftrightarrow A< \dfrac{1}{5}.5+\dfrac{1}{10}.8\)
\(\Leftrightarrow A< 1+\dfrac{4}{5}=\dfrac{9}{5}< 2\)
\(\Leftrightarrow A< 2\left(đpcm\right)\)
b/ Ta có :
\(\dfrac{1}{11}>\dfrac{1}{30}\)
\(\dfrac{1}{12}>\dfrac{1}{30}\)
...............
\(\dfrac{1}{29}>\dfrac{1}{30}\)
\(\dfrac{1}{30}=\dfrac{1}{30}\)
\(\Leftrightarrow\dfrac{1}{11}+\dfrac{1}{12}+........+\dfrac{1}{30}>\dfrac{1}{30}+\dfrac{1}{30}+.......+\dfrac{1}{30}\)
\(\Leftrightarrow B>\dfrac{1}{30}.20=\dfrac{2}{3}\)
\(\Leftrightarrow B>\dfrac{2}{3}\left(đpcm\right)\)
1-\(\dfrac{1}{12}\)-\(\dfrac{1}{20}\)-\(\dfrac{1}{30}\)-...-\(\dfrac{1}{210}\)
=1-(\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{210}\))
=1-(\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{14.15}\))
=1-(\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{14}\)-\(\dfrac{1}{15}\))
=1-(\(\dfrac{1}{3}\)-\(\dfrac{1}{15}\))
=1-\(\dfrac{4}{15}\)
=\(\dfrac{11}{15}\)
thanks