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Ta có \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
Ta có \(B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=\frac{1}{2}-\frac{1}{7}\)
\(=\frac{5}{14}\)
Ta có \(C=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{1}{6}-\frac{1}{22}\)
\(=\frac{4}{33}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(B=\frac{1}{2}-\frac{1}{7}\)
\(B=\frac{5}{14}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(C=\frac{1}{6}-\frac{1}{22}=\frac{4}{33}\)
Lời giải:
a)
$\frac{2}{12}+\frac{7}{-42}=\frac{1}{6}+\frac{-1}{6}=0$
b)
$\frac{24}{40}+\frac{-14}{35}=\frac{3}{5}+\frac{-2}{5}=\frac{1}{5}$
c)
$\frac{-14}{49}+\frac{12}{30}=\frac{-2}{7}+\frac{2}{5}=\frac{4}{35}$
d)
$\frac{6}{-21}+\frac{-9}{36}=\frac{-2}{7}+\frac{-1}{4}=\frac{-15}{28}$
Mình cũng nghĩ như bạn vậy. Thay vì 30/43, bạn ấy lại ghi là 30/42.
\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
Vậy a = 1; b = 2 ; c = 3 ; d = 4
\(A=\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)
\(A=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}\)
\(A=-1+1+\frac{-1}{5}\)
\(A=\frac{-1}{5}\)
\(B=\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)
\(B=\frac{-1}{3}+\frac{2}{5}+\frac{-2}{3}+\frac{-3}{5}+\frac{1}{5}\)
\(B=\left(\frac{-1}{3}+\frac{-2}{3}\right)+\left(\frac{2}{5}+\frac{-3}{5}+\frac{1}{5}\right)\)
\(B=-1+0\)
\(B=-1\)
xét A và B có :
\(\frac{42}{47}\)<\(\frac{42}{45}\) (1)
theo tính chất bắc cầu ta có ;
\(\frac{37}{51}\)+\(\frac{14}{51}\)=1 ; \(\frac{29}{37}\)+\(\frac{8}{37}\)=1
\(\frac{31}{35}\)+\(\frac{4}{35}\)=1 ; \(\frac{49}{63}\)+\(\frac{14}{63}\)=1
Mà \(\frac{14}{51}\)>\(\frac{14}{63}\)=> \(\frac{37}{51}\)< \(\frac{49}{63}\)(2)
ta lại có : \(\frac{4}{35}\)=\(\frac{8}{70}\)( nhân cả tử và mẫu vs 2 )
mà \(\frac{8}{70}\)<\(\frac{8}{37}\)nên \(\frac{4}{35}\)<\(\frac{8}{37}\)=>\(\frac{29}{37}< \frac{31}{35}\)(3)
Từ (1) ; (2);(3)=>\(\frac{42}{47}+\frac{37}{51}+\frac{29}{37}< \frac{42}{45}+\frac{49}{63}+\frac{31}{35}\)
a, \(A=\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{61.66}\)
\(A=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{61}-\frac{1}{66}\)
\(A=\frac{1}{11}-\frac{1}{66}\)
\(A=\frac{5}{66}\)
b, \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(B=1-\frac{1}{7}\)
\(B=\frac{6}{7}\)
_Học tốt nha_
C\(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{5.6}\)-\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)-\(\frac{1}{8.9}+\frac{1}{9.10}\)
c=\(\frac{1}{1}-\frac{1}{10}\)
c=\(\frac{9}{10}\)
còn a và b rễ lắm mình ko thích làm bài rễ đâu bạn cố chờ lời giải khác nhé!