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\(A=\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)
\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
\(=\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{12}{60}-\dfrac{5}{60}=\dfrac{7}{60}\)
Vậy \(A=\dfrac{7}{60}\)
\(A=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)
\(A=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)
\(A=\dfrac{1}{5}-\dfrac{1}{12}\)
\(A=\dfrac{7}{60}\)
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= \(\dfrac{7}{8}\)
\(A=\dfrac{5}{9}-\dfrac{5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\ =\dfrac{5}{9}+\dfrac{-5}{8}+\dfrac{2}{3}+\dfrac{4}{9}+\dfrac{-3}{8}+\dfrac{1}{3}\\= \left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{-5}{8}+\dfrac{-3}{8}\right)\\ =1+1+\left(-1\right)\\ =2+\left(-1\right)\\ =1\)
1,A=\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+...+\(\dfrac{1}{99.100}\)
1,A= \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-...-\(\dfrac{1}{99}\)+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)
1,A= \(\dfrac{1}{2}\)-\(\dfrac{1}{100}\)
1,A= \(\dfrac{49}{100}\)
(Còn câu 2 mình chưa nghĩ ra. Cho mình hỏi 1 câu, không biết là đề câu 2 có chính xác không?)
đề này do cô giáo lớp mình cho mình cũng đc chữa câu 1 rồi còn câu 2 thì tiếp tục nghĩ
=>(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
Đăng ít thôi.
d) \(D=\dfrac{1}{1.2.3}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+\dfrac{1}{5.6.7}+\dfrac{1}{6.7.8}+\dfrac{1}{7.8.9}+\dfrac{1}{8.9.10}\)
\(\Rightarrow2D=\dfrac{2}{1.2.3}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}+\dfrac{2}{7.8.9}+\dfrac{2}{8.9.10}\)
\(\Rightarrow2D=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{4.5}-\dfrac{1}{5.6}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\)
\(\Rightarrow2D=\dfrac{1}{2.3}-\dfrac{1}{9.10}\)
\(\Rightarrow2D=\dfrac{22}{45}\)
\(\Rightarrow D=\dfrac{11}{45}\)
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 \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\)
1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
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=\(\dfrac{-9}{10}\)