`6/7 . 8/13 +6/7 . 9/13+3/13 . 6/7`

`=6/7 . (8/13+9/13+3/13)`

`=6/7 . 20/13`

`=120/91`

\(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{7}.\dfrac{9}{13}+\dfrac{3}{13}.\dfrac{6}{7}\)

\(=\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{9}{13}+\dfrac{3}{13}\right)\)

\(=\dfrac{6}{7}.\left(\dfrac{8+9+3}{13}\right)\)

\(=\dfrac{6}{7}.\dfrac{20}{13}\)

\(=\dfrac{6.20}{7.13}\)

\(=\dfrac{120}{91}\)

`6/7 . 8/13 + 6/7 . 9/13 + 3/13 . 6/7`

`= 6/7 . ( 8/13 + 9/13 + 3/13)`

`=6/7 . 20/13`

`= 120/91`

\(-\dfrac{5}{7}\times\dfrac{2}{13}+\dfrac{-5}{7}\times\dfrac{3}{13}-\dfrac{5}{7}\times\dfrac{8}{13}\)

\(=-\dfrac{5}{7}\left(\dfrac{2}{13}+\dfrac{3}{13}+\dfrac{8}{13}\right)\)

\(=-\dfrac{5}{7}\times\dfrac{13}{13}\)

\(=-\dfrac{5}{7}\times1=-\dfrac{5}{7}\)

\(-\dfrac{5}{7}.\dfrac{2}{13}+\dfrac{-5}{7}.\dfrac{3}{13}-\dfrac{5}{7}.\dfrac{8}{13}\)

\(=-\dfrac{10}{91}-\dfrac{15}{91}-\dfrac{40}{91}\)

\(=\dfrac{-10-15-40}{91}\)

\(=-\dfrac{65}{91}=-\dfrac{5}{7}\)

\(\dfrac{-5}{7}.\dfrac{2}{13}+\left(\dfrac{-5}{7}\right).\dfrac{3}{13}-\dfrac{5}{7}.\dfrac{8}{13}\)

\(=\dfrac{5}{7}.\left(\dfrac{-2}{13}+\dfrac{-3}{13}-\dfrac{8}{13}\right)\)

\(=\dfrac{5}{7}.\left(-1\right)\)

\(=-\dfrac{5}{7}\)

\(#PaooNqoccc\)

a)12 x 18 + 14 x 3 - 255 : 17=25.58823529411765

b)68 + 42 x 5 - 625 : 25=5182

c) 13+ 21 x 5 - (198 : 11 - 8)=108

d) 1188 - 4 x (27 + 90 + 73 :10 )=

e) 5 + 8 + 11 + 14 +......+ 38 + 41=

f) 51 - 49 + 55 - 53 + 59 - 56 + 63 - 61 + 65=74

có mấy câu mình ko biết,xin lỗi bạn

Ta có :

\(\frac{1}{12}=\frac{1}{12}\)

\(\frac{1}{13}< \frac{1}{12}\)

\(\frac{1}{14}< \frac{1}{12}\)

\(........\)

\(\frac{1}{17}< \frac{1}{12}\)

Cộng vế với vế ta có :

\(\frac{1}{12}+\frac{1}{13}+....+\frac{1}{17}< \frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\)(có 6 số \(\frac{1}{12}\))\(=\frac{6}{12}=\frac{1}{2}\)

Vậy \(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{17}< \frac{1}{2}\)

Bài 4:

Điện trở tương đương :

\(R_{tđ}=\dfrac{U}{V}=\dfrac{12}{0,4}=30\left(\Omega\right)\)

Ta có: \(R_{tđ}=R_1+R_2\Leftrightarrow R_1=R_{tđ}-R_2=30-20=10\left(\Omega\right)\)

A B M N R1 R2 R3

Vì R > R_tđ ⇒ R₁ // R₂ \(\Rightarrow R_{tđ\left(MN\right)}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{30.30}{30+30}=15\left(\Omega\right)\)

Vì R_MN > R_tđ ⇒ R_MN // R₃  \(\Rightarrow R_{tđ\left(AB\right)}=\dfrac{R_{MN}.R_3}{R_{MN}+R_3}=\dfrac{15.30}{15+30}=10\left(\Omega\right)\)

Đó nha, khi qua ko đọc kĩ

Mà cái này có nhiều cách vẽ lắm

\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)

\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)

\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)

\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)

\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)

\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)

\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)

\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)

\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)

\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)

\(A=\frac{907,425}{28561}-\frac{19}{37}\)

\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)

\(A=\frac{-509084,275}{1056757}=-0,04604282...\)

Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!

bai 1

=20.(-5)+23.(-30)

=(-100)+(-690)=-790

có thời gian đăng chữ sao ko đăng bài

bài giải :

a, nếu mỗi toa xe chở  20 tấn hàng thì ta cần số toa xe là: 180: 20 =  9 (toa )

b, nếu mỗi toa chở 30 tấn hàng thì cần số toa xe là : 180 :30 = 6 (toa )

đáp số : a: 9 toa

                      b, 6 toa 

đúng không ???