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theo đề bài ta có:
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=1\)
đặt A=4/5.9+4/9.13+4/13.17+...+4/41.45
=1/5-1/9+1/9-1/13+1/13-1/17+...+1/41-1/45
=1/5-1/45
=8/45
suy ra x+8/45=-37/45
suy ra x=-1
\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)
\(\Leftrightarrow x=\frac{7.45}{21}=15\)
bài này mà cũng đăng dễ ẹc
x + 4/5.9 + 4/9.13 + 4/13.17 + ... + 4/41.45 =-37/45
x+( 1/5 - 1/9 +1/9 - 1/13 + 1/13 - 1/17 + ... + 1/41 - 1/45 )= -37/45
x+ (1/5 - 1/45 )=-37/45
x+8/45 = -37 / 45
x=-37/45 -8/45
x=-45/45=-1
Ta có : \(\frac{7}{x-2005}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x-2005}=\frac{29}{45}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)\)
\(\Rightarrow\frac{7}{x-2005}=\frac{29}{45}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(\Rightarrow\frac{7}{x-2005}=\frac{29}{45}-\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}-\frac{8}{45}=\frac{7}{15}\)
\(\Rightarrow x-2005=15\Rightarrow x=15+2005=2020\)
Vậy x =2020
x + 4/5.9 + 4/9.13 + ... + 4/41.45 = -37/45
<=> x + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/41 - 1/45= -37/45
<=> x + 1/5 - 1/45 = -37/45
<=> x + 9/45 = -36/45
<=>x= -45/45=-1
1, Tính tổng:
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)=\frac{5}{7}\cdot\frac{-7}{11}=\frac{-5}{11}\)
2, Tìm x:
\(x+\frac{5}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\Rightarrow x+\frac{9}{45}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\Rightarrow x=\frac{-37}{45}-\frac{8}{45}=\frac{-45}{45}=-1\)
- Các bài tìm x còn lại bạn cứ theo trình tự thực hiện phép tính mà làm nhé!
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)
\(=\frac{5}{7}\cdot-\frac{7}{11}\)
\(=-\frac{5}{11}\)
Mk sẽ giải từng câu :)
Bài 1 :
Gọi \(ƯCLN\left(2n+2;6n+5\right)=d\)
\(\Rightarrow\hept{\begin{cases}2n+2⋮d\\6n+5⋮d\end{cases}\Rightarrow\hept{\begin{cases}6\left(2n+2\right)⋮d\\2\left(6n+5\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}12n+12⋮d\\12n+10⋮d\end{cases}}}\)
\(\Rightarrow\)\(\left(12n+12\right)-\left(12n+10\right)⋮d\)
\(\Rightarrow\)\(2⋮d\)
\(\Rightarrow\)\(d\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Mà \(6n+5\) không chia hết cho \(2\) và \(-2\) nên \(ƯCLN\left(2n+2;6n+5\right)=\left\{1;-1\right\}\)
Vậy \(\frac{2n+2}{6n+5}\) là phân số tối giản với mọi n
Chúc bạn học tốt ~
1. Gọi d = ƯCLN (2n+2,6n+5)
=>\(\hept{\begin{cases}2n+2\\6n+5\end{cases}}\)chia hết cho d
=>\(\hept{\begin{cases}3.\left(2n+2\right)\\6n+5\end{cases}}\)chia hết cho d
=>\(\hept{\begin{cases}6n+6^{\left(1\right)}\\6n+5^{\left(2\right)}\end{cases}}\)chia hết cho d
Từ (1) và (2) => (6n+6) - (6n+5) chia hết cho d
=> 6n + 6 - 6n - 5 chia hết cho d
=> 1 chia hết cho d
=> d =1
=> ƯCLN (2n+2,6n+5) = 1
Vậy \(\frac{2n+2}{6n+5}\) là phân số tối giản
2. Ta có:
B = 32. (\(\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}+...+\frac{3}{67.70}\))
B = 32. (\(\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{67}-\frac{1}{70}\))
B = 32. (\(\frac{1}{10}-\frac{1}{70}\))
B = 27/35
Vì \(\frac{27}{35}< 1\)
=> B < 1
3. x + \(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{-37}{45}\)
x + ( \(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
x + (\(\frac{1}{5}-\frac{1}{45}\)) = \(\frac{-37}{45}\)
x + \(\frac{8}{45}=\frac{-37}{45}\)
x = \(\frac{-37}{45}-\frac{8}{45}\)
x = -1
\(x+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(\Leftrightarrow x+3\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}.\frac{8}{45}=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{2}{15}=-\frac{37}{45}\)
\(\Leftrightarrow x=-\frac{43}{45}\)