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a) Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)(hai góc kề bù)

\(\Leftrightarrow\widehat{zOy}+140^0=180^0\)

hay \(\widehat{yOz}=40^0\)

Vậy: \(\widehat{yOz}=40^0\)

21 tháng 4 2021

a) Ta có: xOy+yOz=1800xOy+yOz=1800(hai góc kề bù)

              ⇔zOy+1400=1800

     hay yOz=400yOz=400

     Vậy: yOz=400

18 tháng 4 2021

mình ko chép lại đề nhé, sửa 2014 + 2016 thành 2014.2016

\(A=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2014}-\dfrac{1}{2016}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2016}\right)=2\left(\dfrac{2016-2}{6032}\right)=\dfrac{2.2018}{6032}=\dfrac{4036}{6032}=\dfrac{1009}{1508}\)

18 tháng 4 2021

\(A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2014.2016}\)

\(=2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{2014.2016}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2014}-\dfrac{1}{2016}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{2016}\right)\)

\(=2.\dfrac{1007}{2016}=\dfrac{1007}{1008}\)

31 tháng 12 2022

=101-102-103+104-105-106+107-108+109+110
= (101+104+107+109+110)-(102+103+105+106+108)
=701-704
=-3

31 tháng 12 2022

ý lộn
 = 531-524
=-7

31 tháng 8 2020

Ai xong trước và đúng thì mik nhé😇😇😇😇😇🥺🥺😭😫😫

31 tháng 8 2020

A B C D E H

=>1/3x=2/3

hay x=2

17 tháng 3 2022

giúp j bn?

17 tháng 3 2022

lỗi câu hỏi rồi bạn

a: =>5x=-1,2

hay x=-0,24

b: =>4x=2,6

hay x=0,65

giải chi tiết hộ mik nha

17 tháng 4 2023

Mình nghĩ là lần sau bạn tách ra tầm khoảng 4-5 ý rồi đăng thì nó sẽ ngắn hơn ạ;-;. Chứ như này thì dài quá -> không có ai giúp đâu nhé:").

 

\(a,\dfrac{-6}{13}+\dfrac{7}{-13}=-\dfrac{13}{13}=-1\)

\(b,\dfrac{7}{12}+\dfrac{-9}{24}=\dfrac{14}{24}+\dfrac{-9}{24}=\dfrac{5}{24}\)

\(c,\dfrac{1}{5}+\dfrac{-5}{19}+\dfrac{4}{5}+\dfrac{-14}{19}=\left(\dfrac{1}{5}+\dfrac{4}{5}\right)+\left(\dfrac{-5}{19}+\dfrac{-14}{19}\right)=1+\left(-1\right)=0\)

\(d,\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)

\(=1+\left(-1\right)+\dfrac{-5}{7}=0+\dfrac{-5}{7}=\dfrac{-5}{7}\)

\(e,\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}=\left(\dfrac{93}{11}+\dfrac{29}{8}\right)-\dfrac{38}{11}=\left(\dfrac{93}{11}-\dfrac{38}{11}\right)+\dfrac{29}{8}\)

\(=\dfrac{55}{11}+\dfrac{29}{8}=5+\dfrac{29}{8}=\dfrac{40}{8}+\dfrac{29}{8}=\dfrac{69}{8}\)

\(f,\dfrac{-1}{-5}+\dfrac{8}{7}+\dfrac{-6}{13}-1\dfrac{23}{7}-\dfrac{13}{24}=\dfrac{1}{5}+\dfrac{8}{7}+\dfrac{-6}{13}-\dfrac{30}{7}-\dfrac{13}{24}\)

\(=\dfrac{7}{35}+\dfrac{40}{35}+\dfrac{-6}{13}-\dfrac{30}{7}-\dfrac{13}{24}=\dfrac{47}{35}+\dfrac{-6}{13}-\dfrac{30}{7}-\dfrac{13}{24}\)

\(=\dfrac{611}{455}+\dfrac{-210}{455}-\dfrac{30}{7}-\dfrac{13}{24}=\dfrac{401}{455}-\dfrac{30}{7}-\dfrac{13}{24}=\dfrac{401}{455}-\dfrac{1950}{455}-\dfrac{13}{24}\)

\(=\dfrac{-1549}{455}-\dfrac{13}{24}=\dfrac{-37176}{10920}-\dfrac{5915}{10920}=\dfrac{-43091}{10920}=-3,946\) (số to quá bạn ơi;-;)

\(g,0,75+\dfrac{-1}{3}+\dfrac{3}{6^2}-\dfrac{5}{12}=\dfrac{3}{4}+\dfrac{-1}{3}+\dfrac{3}{36}-\dfrac{5}{12}=\dfrac{9}{12}+\dfrac{-4}{12}+\dfrac{1}{12}-\dfrac{5}{12}\)

\(=\dfrac{9+\left(-4\right)+1-5}{12}=\dfrac{1}{12}\)

\(h,\dfrac{-3}{7}\cdot\dfrac{-1}{9}+\dfrac{-7}{18}\cdot\dfrac{-3}{7}+\dfrac{5}{6}\cdot\dfrac{-3}{7}=\dfrac{-3}{7}\cdot\left(\dfrac{-1}{9}+\dfrac{-7}{18}+\dfrac{5}{6}+1\right)=\dfrac{-3}{7}\cdot\dfrac{4}{3}=\dfrac{-12}{21}\)

\(i,\dfrac{-5}{12}\cdot\dfrac{4}{19}+\dfrac{-7}{12}\cdot\dfrac{4}{19}-\dfrac{40}{57}=\dfrac{4}{19}\cdot\left(\dfrac{-5}{12}+\dfrac{-7}{12}\right)-\dfrac{40}{57}=\dfrac{4}{19}\cdot\left(-1\right)-\dfrac{40}{57}\)

\(=-\dfrac{4}{19}-\dfrac{40}{57}=-\dfrac{12}{57}-\dfrac{40}{57}=-\dfrac{52}{57}\)

\(j,\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+...+\dfrac{1}{99}\cdot\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{100}{200}-\dfrac{2}{200}=\dfrac{98}{200}=\dfrac{49}{100}\)

\(k,\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{98.100}\)

\(=\dfrac{1}{2}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{6}+...+\dfrac{1}{98}\cdot\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{100}{200}-\dfrac{2}{200}=\dfrac{98}{100}=\dfrac{49}{100}\)

`@mt`

Hoc24.vn

*Do nhiều ý quá nên bài làm của tớ cũng không tránh khỏi nhầm lẫn, sai sót, có gì bạn tính lại nhé:(, tớ cảm ơn 😭.

a: =-6/13-7/13=-13/13=-1

b: =14/24-9/24=5/24

c: =1/5+4/5-5/19-14/19=1-1=0

d: \(=\dfrac{5}{13}+\dfrac{8}{13}-\dfrac{20}{41}-\dfrac{21}{41}-\dfrac{5}{7}=-\dfrac{5}{7}\)

e: \(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}=8+\dfrac{5}{8}=\dfrac{69}{8}\)