(7|6+5|12)x12
=.....x......+......x......
=......+......
=
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(-1,2+\dfrac{2}{3}+x=5\)
=>\(x=5+1,2-\dfrac{2}{3}=6,2-\dfrac{2}{3}\)
=>\(x=\dfrac{31}{5}-\dfrac{2}{3}=\dfrac{93}{15}-\dfrac{10}{15}=\dfrac{83}{15}\)
b: \(2\dfrac{4}{7}-3x=\dfrac{-4}{5}+\dfrac{2}{3}\)
=>\(\dfrac{18}{7}-3x=\dfrac{-12}{15}+\dfrac{10}{15}=\dfrac{-2}{15}\)
=>\(3x=\dfrac{18}{7}+\dfrac{2}{15}=\dfrac{270}{105}+\dfrac{14}{105}=\dfrac{284}{105}\)
=>\(x=\dfrac{284}{315}\)
c: \(\dfrac{1}{6}-\dfrac{3}{8}+1,75=3\dfrac{4}{3}-x\)
=>\(\dfrac{13}{3}-x=\dfrac{4}{24}-\dfrac{9}{24}+\dfrac{42}{24}=\dfrac{37}{24}\)
=>\(x=\dfrac{13}{3}-\dfrac{37}{24}=\dfrac{108}{24}-\dfrac{37}{24}=\dfrac{71}{24}\)
d: \(\dfrac{1}{6}-\dfrac{4}{9}+0,125=2\dfrac{4}{3}-2x\)
=>\(\dfrac{10}{3}-2x=\dfrac{-11}{72}\)
=>\(2x=\dfrac{10}{3}+\dfrac{11}{72}=\dfrac{240}{72}+\dfrac{11}{72}=\dfrac{251}{72}\)
=>\(x=\dfrac{251}{144}\)
e: \(2\dfrac{2}{3}-4x=\dfrac{-7}{5}+\dfrac{2}{3}\)
=>\(2+\dfrac{2}{3}-4x=\dfrac{-7}{5}+\dfrac{2}{3}\)
=>\(2-4x=-\dfrac{7}{5}\)
=>\(4x=2+\dfrac{7}{5}=\dfrac{17}{5}\)
=>\(x=\dfrac{17}{20}\)
f: \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
=>\(x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}=\dfrac{3}{6}-\dfrac{5}{6}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
=>\(x=-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
g: \(\left(\dfrac{3}{5}-\dfrac{4}{3}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{9}{7}\)
=>\(\dfrac{-11}{15}+\dfrac{5}{8}-x=\dfrac{9}{7}\)
=>\(\dfrac{-13}{120}-x=\dfrac{9}{7}\)
=>\(x=-\dfrac{13}{120}-\dfrac{9}{7}=\dfrac{-1171}{840}\)
a, \(-1,2+\dfrac{2}{3}+x=5\Leftrightarrow x=5+1,2-\dfrac{2}{3}=\dfrac{83}{15}\)
b, \(2\dfrac{4}{7}-3x=-\dfrac{4}{5}+\dfrac{2}{3}\Leftrightarrow\dfrac{18}{7}-3x=-\dfrac{2}{15}\Leftrightarrow3x=\dfrac{284}{105}\Leftrightarrow x=\dfrac{284}{315}\)
c, \(\dfrac{1}{6}-\dfrac{3}{8}+1,75=3\dfrac{4}{3}-x\Leftrightarrow-x+\dfrac{13}{3}=\dfrac{37}{24}\Leftrightarrow x=\dfrac{13}{3}-\dfrac{37}{24}=\dfrac{67}{24}\)
d, \(\dfrac{1}{6}-\dfrac{4}{9}+0,125=2\dfrac{4}{3}-2x\Leftrightarrow-2x+\dfrac{10}{3}=-\dfrac{-11}{72}\Leftrightarrow2x=\dfrac{251}{72}\Leftrightarrow x=\dfrac{251}{144}\)
e, \(2\dfrac{2}{3}-4x=-\dfrac{7}{5}+\dfrac{2}{7}\Leftrightarrow\dfrac{8}{3}-4x=-\dfrac{39}{35}\Leftrightarrow4x=\dfrac{397}{105}\Leftrightarrow x=\dfrac{397}{420}\)
f, \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{2}{3}\)
g, \(\left(\dfrac{3}{5}-\dfrac{4}{3}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{9}{7}\Leftrightarrow\dfrac{-11}{15}+\dfrac{5}{8}-x=\dfrac{9}{7}\Leftrightarrow\left(-\dfrac{13}{120}\right)-x=\dfrac{9}{7}\Leftrightarrow x=-\dfrac{1171}{840}\)
\(\left(\dfrac{7}{6}+\dfrac{5}{12}\right)\times12\)
\(=\left(\dfrac{14}{12}+\dfrac{5}{12}\right)\times12\)
\(=\dfrac{19}{12}\times12=19\)
TK ạ
Nếu chỉ có hai lực tác dụng vào cùng một vật mà vật vẫn đứng yên thì hai lực đó là hai lực cân bằng.
Ví dụ: Hai đội kéo co cùng kéo sợi dây. Nếu hai đội mạnh ngang nhau thì họ sẽ tác dụng lên dây hai lực cân bằng. Sợi dây chịu tác dụng của hai lực cân bằng thì sẽ đứng yên. Hai vecto→u𝑢→ và →v𝑣→ biểu diễn cho hai vecto cân bằng thì hai vecto này có chung gốc, ngược hướng và có độ lớn (hay độ dài) bằng nhau.
Hai lực cân bằng là hai lực có cùng phương, ngược chiều, cùng độ lớn và có cùng điểm đặt (tác động vào cùng một điểm).
Nếu biểu diễn bằng vector thì 2 vector này cùng phương, ngược chiều, có độ dài bằng nhau và có chung điểm gốc.
a: \(\dfrac{-3}{5}+\dfrac{28}{5}\left(\dfrac{13}{56}-\dfrac{5}{24}+\dfrac{1}{7}\right)\)
\(=\dfrac{-3}{5}+\dfrac{28}{5}\left(\dfrac{39}{168}-\dfrac{35}{168}+\dfrac{24}{168}\right)\)
\(=\dfrac{-3}{5}+\dfrac{28}{5}\cdot\dfrac{28}{168}=\dfrac{-3}{5}+\dfrac{28}{5}\cdot\dfrac{1}{6}=\dfrac{-3}{5}+\dfrac{14}{15}\)
\(=\dfrac{-9}{15}+\dfrac{14}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
b: \(\dfrac{5}{7}\cdot\dfrac{11}{18}+\dfrac{3}{7}\cdot\dfrac{5}{18}+\dfrac{4}{9}\)
\(=\dfrac{5}{18}\left(\dfrac{11}{7}+\dfrac{3}{7}\right)+\dfrac{4}{9}\)
\(=\dfrac{5}{18}\cdot2+\dfrac{4}{9}=\dfrac{5}{9}+\dfrac{4}{9}=\dfrac{9}{9}=1\)
c: \(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)
\(=\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\dfrac{-5}{7}=10\cdot\dfrac{-7}{5}=-14\)
d: \(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)
\(=\left(-\dfrac{3}{5}+\dfrac{4}{9}+\dfrac{-2}{5}+\dfrac{5}{9}\right)\cdot\dfrac{11}{7}\)
\(=\left(-\dfrac{5}{5}+\dfrac{9}{9}\right)\cdot\dfrac{11}{7}=\left(-1+1\right)\cdot\dfrac{11}{7}=0\)
e: \(\dfrac{-3}{4}\cdot5\dfrac{3}{13}-0,75\cdot\dfrac{36}{13}\)
\(=\dfrac{-3}{4}\left(5+\dfrac{3}{13}+\dfrac{36}{13}\right)\)
\(=\dfrac{-3}{4}\cdot8=-6\)
f: \(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{302\cdot305}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{302\cdot305}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{302}-\dfrac{1}{305}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{305}\right)=\dfrac{1}{3}\cdot\dfrac{12}{61}=\dfrac{4}{61}\)
g: \(6\dfrac{5}{12}:2\dfrac{3}{4}-11\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}-\dfrac{45}{4}\cdot\dfrac{2}{15}=\dfrac{77}{12}\cdot\dfrac{4}{11}-\dfrac{3}{2}\)
\(=\dfrac{7}{3}-\dfrac{3}{2}=\dfrac{14}{6}-\dfrac{9}{6}=\dfrac{5}{6}\)
h: \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right)\cdot2\dfrac{2}{3}\cdot0,25\)
\(=\left(0,6+0,415-0,015\right)\cdot\dfrac{8}{3}\cdot\dfrac{1}{4}\)
\(=1\cdot\dfrac{2}{3}=\dfrac{2}{3}\)
i: \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right)\cdot\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right)\cdot\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right)\cdot\dfrac{10}{11}=\dfrac{5}{2}-\dfrac{33}{20}\cdot\dfrac{10}{11}=\dfrac{5}{2}-\dfrac{3}{2}=\dfrac{2}{2}=1\)
\(C=\left(6-\dfrac{2}{3}+\dfrac{5}{16}\right)-\left(2+\dfrac{4}{3}-\dfrac{7}{8}\right)-\left(6-\dfrac{5}{8}+\dfrac{13}{12}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{5}{16}-2-\dfrac{4}{3}+\dfrac{7}{8}-6+\dfrac{5}{8}-\dfrac{13}{12}\)
\(=\left(6-2-6\right)+\left(-\dfrac{2}{3}-\dfrac{4}{3}\right)+\left(\dfrac{5}{16}+\dfrac{7}{8}+\dfrac{5}{8}\right)-\dfrac{13}{12}\)
\(=-2-2+\left(\dfrac{5}{16}+\dfrac{12}{8}\right)-\dfrac{13}{12}\)
\(=-4-\dfrac{13}{12}+\dfrac{29}{16}=-\dfrac{157}{48}\)
\(C=\left(6-\dfrac{2}{3}-\dfrac{5}{10}\right)-\left(2+\dfrac{4}{3}-\dfrac{7}{8}\right)-\left(6-\dfrac{5}{8}+\dfrac{13}{12}\right)\)
\(=6-\dfrac{2}{3}-\dfrac{5}{10}-2-\dfrac{4}{3}+\dfrac{7}{8}-6+\dfrac{5}{8}-\dfrac{13}{12}\)
\(=\left(6-2-6\right)+\left(-\dfrac{2}{3}-\dfrac{4}{3}\right)+\left(\dfrac{7}{8}+\dfrac{5}{8}\right)+\left(\dfrac{-5}{10}-\dfrac{13}{12}\right)\)
\(=\left(-2\right)+\left(-2\right)+\dfrac{3}{2}+\dfrac{-19}{12}\)
\(=\left(-4\right)+\dfrac{18}{12}+\dfrac{-19}{12}\)
\(=\left(-4\right)+\dfrac{-1}{12}\)
\(=\dfrac{-48}{12}+\dfrac{-1}{12}\)
\(=\dfrac{-49}{12}\)
\(B=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{4}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{4}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(-\dfrac{3}{4}-\dfrac{7}{4}+\dfrac{5}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)\)
\(=-3+\dfrac{-5}{4}+\dfrac{-7}{5}=-3-1,25-1,4=-5,65\)
a) \(\dfrac{-3}{100}>\dfrac{-50}{100}=-\dfrac{1}{2}\)
\(\dfrac{-2}{3}< \dfrac{-1,5}{3}=-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{-3}{100}>\dfrac{-2}{3}\)
b) \(\dfrac{-3}{5}=\dfrac{-9}{15}\)
\(\dfrac{-2}{3}=\dfrac{-10}{15}\)
Mà: - 9 > -10
\(\Rightarrow-\dfrac{9}{15}>\dfrac{-10}{15}\)
hay `-3/5>-2/3`
c) \(\dfrac{-5}{4}< \dfrac{-2}{4}=-\dfrac{1}{2}\)
\(-\dfrac{3}{8}>\dfrac{-4}{8}=-\dfrac{1}{2}\)
\(\Rightarrow-\dfrac{5}{4}< \dfrac{-3}{8}\)
d) \(-\dfrac{2}{3}=\dfrac{1}{3}-1\)
\(-\dfrac{3}{4}=\dfrac{1}{4}-1\)
Vì: `1/3>1/4`
`=>1/3-1>1/4-1`
Hay `-2/3>-3/4`
a: \(\dfrac{-3}{100}=\dfrac{-3\cdot3}{100\cdot3}=\dfrac{-9}{300};\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot100}{3\cdot100}=\dfrac{-200}{300}\)
mà -9>-200
nên \(\dfrac{-3}{100}>\dfrac{-2}{3}\)
b: \(\dfrac{-3}{5}=\dfrac{-3\cdot3}{5\cdot3}=\dfrac{-9}{15};\dfrac{2}{-3}=\dfrac{-2}{3}=\dfrac{-2\cdot5}{3\cdot5}=\dfrac{-10}{15}\)
mà -9>-10
nên \(\dfrac{-3}{5}>\dfrac{2}{-3}\)
c: \(\dfrac{-5}{4}=\dfrac{-5\cdot2}{4\cdot2}=\dfrac{-10}{8};\dfrac{-3}{8}=\dfrac{-3}{8}\)
mà -10<-3
nên \(-\dfrac{5}{4}< -\dfrac{3}{8}\)
d: \(\dfrac{-2}{3}=\dfrac{-2\cdot4}{3\cdot4}=\dfrac{-8}{12};\dfrac{3}{-4}=\dfrac{-3}{4}=\dfrac{-3\cdot3}{4\cdot3}=\dfrac{-9}{12}\)
mà -8>-9
nên \(-\dfrac{2}{3}>\dfrac{3}{-4}\)
e: \(\dfrac{267}{-268}=\dfrac{-267}{268}>-1;-1=\dfrac{-1343}{1343}>\dfrac{-1347}{1343}\)
Do đó: \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)
f: \(\dfrac{2022\cdot2023-1}{2022\cdot2023}=1-\dfrac{1}{2022\cdot2023}\)
\(\dfrac{2023\cdot2024-1}{2023\cdot2024}=1-\dfrac{1}{2023\cdot2024}\)
Ta có: 2022<2024
=>\(2022\cdot2023< 2023\cdot2024\)
=>\(\dfrac{1}{2022\cdot2023}>\dfrac{1}{2023\cdot2024}\)
=>\(-\dfrac{1}{2022\cdot2023}< -\dfrac{1}{2023\cdot2024}\)
=>\(\dfrac{-1}{2022\cdot2023}+1< \dfrac{-1}{2023\cdot2024}+1\)
=>\(\dfrac{2022\cdot2023-1}{2022\cdot2023}< \dfrac{2023\cdot2024-1}{2023\cdot2024}\)
g: \(\dfrac{2022\cdot2023}{2022\cdot2023+1}=1-\dfrac{1}{2022\cdot2023+1}\)
\(\dfrac{2023\cdot2024}{2023\cdot2024+1}=1-\dfrac{1}{2023\cdot2024+1}\)
Vì \(2022\cdot2023+1< 2023\cdot2024+1\)
nên \(\dfrac{1}{2022\cdot2023+1}>\dfrac{1}{2023\cdot2024+1}\)
=>\(\dfrac{-1}{2022\cdot2023+1}< \dfrac{-1}{2023\cdot2024+1}\)
=>\(\dfrac{-1}{2022\cdot2023+1}+1< \dfrac{-1}{2023\cdot2024}+1\)
=>\(\dfrac{2022\cdot2023}{2022\cdot2023+1}< \dfrac{2023\cdot2024}{2023\cdot2024+1}\)
\(\left(\dfrac{7}{6}+\dfrac{5}{12}\right)\times12\)
\(=\dfrac{7}{6}\times12+\dfrac{5}{12}\times12\)
\(=14+5\)
\(=19\)