\(A=\dfrac{121212}{131313}+\dfrac{154154}{143143}-\left(\dfrac{2121}{2323}+\dfrac{275275}{253253}\right)\)

\(=\dfrac{12}{13}+\dfrac{14}{13}-\left(\dfrac{21}{23}+\dfrac{25}{23}\right)\)

\(=2-2=0\)

giúp nhanh với 

\(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\Leftrightarrow\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(\Rightarrow\left(\frac{21}{23}+\frac{-21}{23}\right)+\left(\frac{125}{93}+\frac{-125}{143}\right)=\frac{17000}{3999}\)

\(\frac{4}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}=\frac{2}{23}+\frac{125}{93}-\frac{21}{23}-\frac{125}{143}\)

\(=-0,36\)

Nhơ nha bạn.Tích mình mình tích lại

a) A= ( 136/15 - 28/5 + 62/10) .21/24

A= (136/15 - 84/15 + 93/15) . 7/8

A= 145/15. 7/8

A= 203/24

b) B= 70 .( 131313.565656 + 131313/727272 + 131313/ 909090)

= 70. ( 13/56+ 13/72 + 13/90)

= 70 . 39/70

= 39

Câu hỏi của ✨♔♕ Saiko ♕♔✨ - Toán lớp 6 - Học toán với OnlineMath

a) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

= \(\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=\frac{0}{5}+\frac{0}{21}+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

 b) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)

\(=0+\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{17875}{13299}+\frac{-11625}{13299}\)

\(=\frac{6250}{13299}\)

(Hình như câu b hơi sai)

\(a)\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{10}{21}+\frac{3}{20}\)

\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)

\(=\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)

\(=0+0+\frac{3}{20}\)

\(=\frac{3}{20}\)

\(b)\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)

\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)

\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)

\(=0+\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{125}{93}+\frac{-125}{143}\)

\(=\frac{6250}{13299}\)

Sao câu b) kq lại lớn vậy nhỉ , đề bài sai ? , hay mình làm sai ?

\(B=70\cdot\left(\frac{131313}{565656}+\frac{131313}{727272}+\frac{131313}{909090}\right)\)

\(B=70\cdot\left(\frac{13}{56}+\frac{13}{72}+\frac{13}{90}\right)\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\right]\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\right]\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\right]\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{7}-\frac{1}{10}\right)\right]\)

\(B=70\cdot13\cdot\frac{3}{70}\)

\(B=70\cdot\frac{3}{70}\cdot13\)

\(B=3\cdot13\)

\(B=39\)

a) (-1)^a =1 với a chẵn, (-1)^a =-1 với a lẻ

\(A=\left(-1\right)^{1+2+3+4+..+2010+2011}=\left(-1\right)^{\frac{2011+1}{2}.2011}=\left(-1\right)^{1006.2011}=1\)

Vì 1006 là số chẵn => 1006.2011 là số chẵn

b) \(B=70.\left(\frac{13.10101}{56.10101}+\frac{13.10101}{72.10101}+\frac{13.10101}{90.10101}\right)=70.\left(\frac{13}{56}+\frac{13}{72}+\frac{13}{90}\right)=3.13=39\)

c) Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)

=> C=4

\(A=\left(-1\right)\cdot\left(-1\right)^2\cdot\left(-1\right)^3\cdot\left(-1\right)^4\cdot...\cdot\left(-1\right)^{2010}\cdot\left(-1\right)^{2011}\)

\(A=\left(-1\right)\cdot1\cdot\left(-1\right)\cdot1\cdot...\cdot1\cdot\left(-1\right)\)                \(\left(\text{Có 1006 số hạng }\left(-1\right)\right)\left(\text{Chẵn}\right)\) \(\Rightarrow\text{ }\left(-1\right)\cdot\left(-1\right)\cdot\left(-1\right)\cdot...\cdot\left(-1\right)=1\)

\(A=1\cdot1\cdot1\cdot...\cdot1\)

\(A=1\)