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Bài 1: Tìm \(x\)
a; \(x-2\) + 7 = 1.3.(-9)
\(x\) - 2 + 7 = 3.(-9)
\(x\) - 2 + 7 = - 27
\(x\) = - 27 - 7 + 2
\(x\) = - 34 + 2
\(x\) = - 32
Vậy \(x=-32\)
Bài 1
c; - 2\(x\) + 5 = 7
- 2\(x\) = 7 - 5
- 2\(x\) = - 2
\(x\) = -2 : (-2)
\(x\) = - 1
Vậy \(x\) = - 1
Bài 1:
a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)
b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)
c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)
d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)
hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)
e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)
hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)
a) \(x-\dfrac{3}{4}=6\times\dfrac{3}{8}\)
\(x-\dfrac{3}{4}=\dfrac{9}{4}\)
=> \(x=\dfrac{9}{4}+\dfrac{3}{4}=3\)
b) \(\dfrac{7}{8}:x=3-\dfrac{1}{2}\)
\(\dfrac{7}{8}:x=\dfrac{5}{2}\)
=> \(x=\dfrac{7}{8}:\dfrac{5}{2}=\dfrac{7}{20}\)
c) \(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)
\(x+\dfrac{1}{6}=\dfrac{3}{4}\)
=> \(x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)
d) \(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)
\(\dfrac{6}{5}-x=\dfrac{2}{3}\)
=> \(x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{8}{15}\)
e) \(x\times3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\)(?)
\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)
=> \(x=\dfrac{40}{51}:\dfrac{10}{3}=\dfrac{4}{17}\)
f) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\)
\(\dfrac{17}{3}:x=\dfrac{5}{3}\)
=> \(x=\dfrac{17}{3}:\dfrac{5}{3}=\dfrac{17}{5}\)
a: =>x-3/4=18/8=9/4
=>x=9/4+3/4=12/4=3
b: =>7/8:x=5/2
=>x=7/8:5/2=7/8*2/5=14/40=7/20
c: x+1/2*1/3=3/4
=>x+1/6=3/4
=>x=3/4-1/6=9/12-2/12=7/12
d: =>12/10-x=2/3
=>6/5-x=2/3
=>x=6/5-2/3=18/15-10/15=8/15
e: =>x*10/3=10/3:17/4=10/3*4/17
=>x=4/17
f: =>17/3:x=13/3-5/2=26/6-15/6=11/6
=>x=17/3:11/6=17/3*6/11=34/11
Bài 2:
a: \(\dfrac{7}{8}+x=\dfrac{3}{5}\)
=>\(x=\dfrac{3}{5}-\dfrac{7}{8}=\dfrac{24-35}{40}=\dfrac{-11}{40}\)
b: \(\dfrac{17}{2}:x=5\)
=>\(x=\dfrac{17}{2}:5\)
=>\(x=\dfrac{17}{2\cdot5}=\dfrac{17}{10}\)
c: \(x-\dfrac{3}{8}=2+\dfrac{1}{4}\)
=>\(x-\dfrac{3}{8}=\dfrac{9}{4}\)
=>\(x=\dfrac{9}{4}+\dfrac{3}{8}=\dfrac{18}{8}+\dfrac{3}{8}=\dfrac{21}{8}\)
d: \(\dfrac{1}{2}+\dfrac{3}{5}\left(x-2\right)=\dfrac{1}{5}\)
=>\(\dfrac{3}{5}\left(x-2\right)=\dfrac{1}{5}-\dfrac{1}{2}=\dfrac{-3}{10}\)
=>\(x-2=-\dfrac{1}{2}\)
=>\(x=2-\dfrac{1}{2}=\dfrac{3}{2}\)
Bài 1:
a: \(\dfrac{-3}{4}+\dfrac{1}{5}=\dfrac{-15}{20}+\dfrac{4}{20}=\dfrac{-15+4}{20}=\dfrac{-11}{20}\)
b: \(\dfrac{-2}{5}-\dfrac{1}{3}=\dfrac{-6}{15}-\dfrac{5}{15}=\dfrac{-6-5}{15}=\dfrac{-11}{15}\)
c: \(\dfrac{3}{7}\cdot\dfrac{2}{5}-\dfrac{2}{5}=\dfrac{2}{5}\left(\dfrac{3}{7}-1\right)=\dfrac{2}{5}\cdot\dfrac{-4}{7}=\dfrac{-8}{35}\)
d: \(\dfrac{1}{4}+\dfrac{3}{4}\left(\dfrac{2}{3}-\dfrac{1}{2}\right)\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{4-3}{6}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\cdot\dfrac{1}{6}=\dfrac{1}{4}+\dfrac{1}{8}=\dfrac{3}{8}\)
e: \(\dfrac{7}{2}\cdot\dfrac{8}{13}+\dfrac{8}{13}\cdot\dfrac{-5}{12}+\dfrac{8}{13}\)
\(=\dfrac{8}{13}\left(\dfrac{7}{2}-\dfrac{5}{2}+1\right)\)
\(=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)
f: \(1+\dfrac{1}{8}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{80}+\dfrac{1}{120}\)
\(=1+\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}+\dfrac{1}{10\cdot12}\)
\(=1+\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{10\cdot12}\right)\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{12}\right)\)
\(=1+\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{12}\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{5}{12}=1+\dfrac{5}{24}=\dfrac{29}{24}\)