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a)\(\dfrac{7}{21}\) + \(\dfrac{9}{-36}\) = \(\dfrac{7}{21}\)+\(\dfrac{-9}{36}\)=\(\dfrac{1}{3}\)+\(\dfrac{-1}{4}\)=\(\dfrac{4}{12}\)+\(\dfrac{-3}{12}\)=\(\dfrac{1}{12}\)
b) \(\dfrac{-12}{18}\)+\(\dfrac{-21}{35}\)=\(\dfrac{-2}{3}\)+\(\dfrac{-3}{5}\)=\(\dfrac{-10}{15}\)+\(\dfrac{-9}{15}\)=\(\dfrac{-19}{15}\)
c) \(\dfrac{-3}{21}\)+\(\dfrac{6}{42}\)=\(\dfrac{-1}{7}\)+\(\dfrac{1}{7}\)=0
d) \(\dfrac{-18}{24}\)+\(\dfrac{15}{-21}\)=\(\dfrac{-18}{24}\)+\(\dfrac{-15}{21}\)=\(\dfrac{-3}{4}\)+\(\dfrac{-5}{7}\)=\(\dfrac{-21}{28}\)+\(\dfrac{-20}{28}\)=\(\dfrac{-41}{28}\)
a: \(y:\dfrac{2}{3}=-\dfrac{7}{2}+\dfrac{3}{5}=\dfrac{-35+6}{10}=\dfrac{-29}{10}\)
nên \(y=-\dfrac{29}{10}\cdot\dfrac{2}{3}=\dfrac{-58}{30}=-\dfrac{29}{15}\)
b: \(\dfrac{y}{-7}=-\dfrac{20}{35}\)
nên y/7=20/35
hay y=4
c: \(\dfrac{4}{5}=\dfrac{36}{-y}\)
nên \(y=-36\cdot\dfrac{5}{4}=-45\)
\(2\dfrac{4}{5}x-50:\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50:\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51.\dfrac{2}{3}\)
\(\dfrac{14}{5}x-50=34\)
\(\dfrac{14}{5}x=34+50\)
\(\dfrac{14}{5}x=84\)
\(x=84:\dfrac{14}{5}\)
\(x=84.\dfrac{5}{14}\)
\(x=30\)
\(2\dfrac{4}{5}x-50\div\dfrac{2}{3}=51\)
=>\(2\dfrac{4}{5}x-50=51\times\dfrac{2}{3}\)
\(2\dfrac{4}{5}x-50=34\)
=>\(2\dfrac{4}{5}x=34+50\)
\(2\dfrac{4}{5}x=84\)
=>\(x=84\div2\dfrac{4}{5}\)
\(x=30\)
Vậy x=30
a)\(\dfrac{-3}{29}+\dfrac{16}{58}\)\(=\dfrac{-3}{29}+\dfrac{8}{29}=\dfrac{5}{29}\)
b) \(\dfrac{8}{40}+\dfrac{-36}{45}=\dfrac{1}{5}+\dfrac{-4}{5}=\dfrac{-3}{5}\)
c) \(\dfrac{-8}{18}+\dfrac{-15}{27}=\dfrac{-4}{9}+\dfrac{-5}{9}=\dfrac{-9}{9}=-1\)
a) \(\dfrac{-3}{29}+\dfrac{16}{58}=\dfrac{-3}{29}+\dfrac{8}{29}=\dfrac{-3+8}{29}=\dfrac{5}{29}\)
b) \(\dfrac{8}{40}+\dfrac{-36}{45}=\dfrac{1}{5}+\dfrac{-4}{5}=\dfrac{1+\left(-4\right)}{5}=\dfrac{-3}{5}\)
c) \(\dfrac{-8}{18}+\dfrac{-15}{27}=\dfrac{-4}{9}+\dfrac{-5}{9}=\dfrac{-4+\left(-5\right)}{9}=\dfrac{-9}{9}=-1\)
a) \(\dfrac{2}{3}.x-\dfrac{1}{2}.x=\dfrac{5}{12}\)
=> \(\left(\dfrac{2}{3}-\dfrac{1}{2}\right).x=\dfrac{5}{12}\)
=> \(\left(\dfrac{4}{6}-\dfrac{3}{6}\right).x=\dfrac{5}{12}\)
=> \(\dfrac{1}{6}\) . x = \(\dfrac{5}{12}\)
=> \(x=\dfrac{5}{12}:\dfrac{1}{6}\)
=> x =\(\dfrac{5}{12}.\dfrac{6}{1}\)
=> x = \(\dfrac{5}{2}\)
Vậy x = \(\dfrac{5}{2}\)
a. \(\dfrac{-3}{5}\)
b. \(\dfrac{-2}{3}\) c. \(\dfrac{4}{39}\) d. \(\dfrac{26}{45}\)
3. Gọi d là ƯCLN(2n + 3, 4n + 8), d ∈ N*
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+8⋮d\end{cases}\Rightarrow\hept{\begin{cases}2\left(2n+3\right)⋮d\\4n+8⋮d\end{cases}\Rightarrow}\hept{\begin{cases}4n+6⋮d\\4n+8⋮d\end{cases}}}\)
\(\Rightarrow\left(4n+8\right)-\left(4n+6\right)⋮d\)
\(\Rightarrow2⋮d\)
\(\Rightarrow d\in\left\{1;2\right\}\)
Mà 2n + 3 không chia hết cho 2
\(\Rightarrow d=1\)
\(\RightarrowƯCLN\left(2n+3,4n+8\right)=1\)
\(\Rightarrow\frac{2n+3}{4n+8}\) là phân số tối giản.
(2⁵⁰.5³⁵)/(2⁵¹.5³⁶)
= (2⁵⁰/2⁵¹).(5³⁵/5³⁶)
= 1/2 . 1/5
= 1/10
\(\dfrac{2^{50}\cdot5^{35}}{2^{51}\cdot5^{36}}\)
\(=\dfrac{2^{50}\cdot5^{35}}{2\cdot2^{50}\cdot5^{35}\cdot5}\)
\(=\dfrac{1}{2\cdot5}\)
\(=\dfrac{1}{10}\)