Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{-3}{5}+\frac{28.43}{5.56}+\frac{28.5}{5.24}-\frac{28.21}{5.63}\)
=\(\frac{-3}{5}+\frac{2.2.7.43}{5.2.2.7.2}+\frac{2.2.7.5}{2.2.2.3.5}-\frac{2.2.7.3.7}{5.7.3.3}\)
=\(\frac{-3}{5}+\frac{43}{5.2}+\frac{7}{2.3}-\frac{28}{15}\)
=\(\frac{-3}{5}+\frac{43}{10}+\frac{7}{6}-\frac{28}{15}\)
=\(\frac{-18}{30}+\frac{129}{30}+\frac{35}{30}-\frac{56}{30}\)
=\(\frac{-18+129+35-56}{30}\)
=\(\frac{90}{30}=3\)
Chúc bn học tốt
b
.ta có
=-3/5+28/5.43/56+28/5.5/24-28/5.21/63
=-3/5+28/5.(43/56+5/24-21/63)
=5.9/14=45/14
suy ra
biểu thức b có giá trị là 45/14
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
\(A=\frac{5}{13}+\frac{-5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(\Leftrightarrow A=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=1+\left(-1\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=0+\frac{-5}{7}=\frac{-5}{7}\)
Vậy A = \(\frac{-5}{7}\)
B= \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
\(\Leftrightarrow B=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\Leftrightarrow B=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
\(\Leftrightarrow B=-1+1+\frac{-2}{11}\)
\(\Leftrightarrow B=0+\frac{-2}{11}\)
\(\Leftrightarrow\) \(B=\frac{-2}{11}\)
Vậy \(B=\frac{-2}{11}\)
@@ Học tốt
Chiyuki Fujito
K cần tk nhá
Ta có : \(A=\frac{1}{1.101}+\frac{1}{2.202}+\frac{1}{3.103}+...+\frac{1}{10.110}\)
=\(\frac{1}{100}.\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right)\)
= \(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right)\)
= \(\frac{1}{100}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)
Lại có : B = \(\frac{1}{10}.\left(\frac{10}{1.11}+\frac{10}{2.12}+\frac{10}{3.13}+...+\frac{10}{100.110}\right)\)
= \(\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{110}\right)\)
= \(\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)
Khi đó \(A:B=\frac{A}{B}=\frac{\frac{1}{100}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}{\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}=\frac{1}{10}\)
-3/5 + 28.43/5.56 + 28.5/5.24 - 28.21/5.63
=-3/5 + 28.43/5.28.2 + 4.7.5/5.4.6 - 28.21/5.3.21
=-3/5 + 43/10 + 7/6 - 28/15
=-18/30 + 129/30 + 35/30 - 56/30
=-12+129+35-56/30
=94/30=47/15