Kết luận

\(85\left(-51\right)+34\left(-51\right)-19\left(-51\right)\)

\(=\left(-51\right)\left(85+34-19\right)\)

\(=-51\cdot100=-5100\)

    \(85\cdot\left(-51\right)+34\cdot\left(-51\right)-19\cdot\left(-51\right)\)
⇔\(\left[-51\right]\left(85+34-19\right)\)
=  \(\left[-51\right]\cdot100\)
\(=-5100\)

C. 51; 68 nha bn

\(B\left(17\right)\in\left\{17;34;51;68;85;...\right\}\)

Mà B(17) >50, <70

\(\Rightarrow B\left(17\right)\in\left\{51;68\right\}\)

Chọn C

\(\frac{85-17+34}{51-102}=\frac{102}{-51}=-2\)

Ta thấy:
\(\frac{85-17+34}{51-102}=\frac{17\cdot5-17+17\cdot2}{17\cdot3-17\cdot6}=\frac{17\cdot\left(5-1+2\right)}{17\cdot\left(3-6\right)}=\frac{17\cdot\left(4+2\right)}{17\cdot\left[3+\left(-6\right)\right]}=\frac{17\cdot6}{17\cdot-3}=-2\)

bạn viết sai đề rồi phải là 26/43.49 mới đúng rồi làm thế này

A=34/7.13+51/13.22+85/22.37+68/37.49

=17(2/7.13+3/13.22+5/22.37+4/37.49)

=17/3(6/7.13+9/13.22+15/22.37+12/37.49)

=17/3(1/7-1/13+1/13-1/22+1/22-1/37+1/37-1/49)=17/3(1/7-1/49)

b=13(3/7.16+5/16.31+4/31.43+2/43.49)

b=13/3(9/7.16+15/16.31+12/31.43+6/43.49)

=13/3(1/7-1/16+1/16-1/31+1/31-1/43+1/43-1/49)=13/3(1/7-1/49)

=>A/B=17/3/13/3=17/13

chuẩn CMNR

\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)

\(=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)

\(=\frac{17}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)

\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)

\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)

\(=13\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)

\(=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)

\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)

\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}{\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)

1. Ta có : (\(\dfrac{-3}{8}\))³ < 0
(\(\dfrac{8}{243}\))³ > 0
=> (\(\dfrac{-3}{8}\))³ < (\(\dfrac{8}{243}\))³
@Cuber Việt

\(\left(\dfrac{-3}{8}\right)^3< 0< \left(\dfrac{8}{243}\right)^3\)

Vậy \(\left(\dfrac{-3}{8}\right)^3< \left(\dfrac{8}{243}\right)^3\)

\(A=\dfrac{34}{7\cdot13}+\dfrac{51}{13\cdot22}+\dfrac{85}{22\cdot37}+\dfrac{68}{37\cdot49}\\ =\dfrac{17}{3}\cdot\dfrac{6}{7\cdot13}+\dfrac{17}{3}\cdot\dfrac{9}{13\cdot22}+\dfrac{17}{3}\cdot\dfrac{15}{22\cdot37}+\dfrac{17}{3}\cdot\dfrac{12}{37\cdot49}\\ =\dfrac{17}{3}\cdot\left(\dfrac{6}{7\cdot13}+\dfrac{9}{13\cdot22}+\dfrac{15}{22\cdot37}+\dfrac{12}{37\cdot49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\dfrac{6}{49}\\ =\dfrac{34}{49}\)