Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=--\frac{37}{45}.\)
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{37}{45}\)
\(x+\frac{1}{5}-\frac{1}{45}=\frac{37}{45}\)
\(x+\frac{1}{5}=\frac{37}{45}+\frac{1}{45}=\frac{38}{45}\)
\(x=\frac{38}{45}-\frac{1}{5}=\frac{29}{45}\)
\(b,\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2015}{2016}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2015}{2016}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2015}{2016}=\frac{1}{2016}\)
\(\Rightarrow5x+6=2016\)
\(\Rightarrow5x=2010\Rightarrow x=402\)
\(c,\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x\left(x+2\right)}=\frac{2017}{2018}\)
\(\Rightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{2017}{2018}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{2017}{2018}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2017}{2018}=\frac{1}{2018}\)
\(\Rightarrow x+2=2018\Rightarrow x=2016\)
học tốt ~~~
.........................
= \(\frac{1}{2}\). ( \(\frac{2}{1.3}\) + \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) ... + \(\frac{2}{x.\left(x+2\right)}\) )
= \(\frac{1}{2}\) . ( 1 - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{7}\) + ... + \(\frac{1}{x}\)- \(\frac{1}{x+2}\) )
= ................
Bạn tự làm tiếp nhé ! Chúc bạn học tốt :)
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!
\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)
\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)
\(=\frac{5}{7}\cdot-\frac{7}{11}\)
\(=-\frac{5}{11}\)
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 THỂ THẤY, VẾ TRÁI CÓ: 12 CẶP
=> \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
<=> \(x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\) (****)
Ta xét: \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
=> \(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\)
=> \(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\)
=> \(2A=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Ta tiếp tục xét: \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
=> \(3B=1+\frac{1}{3}+...+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)
=> \(2B=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(B=\frac{121}{243}\)
THAY CÁC GIÁ TRỊ A; B VÀO PT (****) TA ĐƯỢC:
=> \(x+\frac{12}{25}=\frac{121}{243}\)
<=> \(x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
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
\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)
\(\Leftrightarrow\)\(100=2x+4\)
\(\Leftrightarrow\)\(2x=96\)
\(\Leftrightarrow\)\(48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
\(a)\) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{47.49}=\frac{24}{x+1}\)
\(\Leftrightarrow\)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{47.49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{47}-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(1-\frac{1}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(\frac{48}{49}=\frac{48}{x+1}\)
\(\Leftrightarrow\)\(49=x+1\)
\(\Leftrightarrow\)\(x=48\)
Vậy \(x=48\)
Chúc bạn học tốt ~
a) \(x+\)\(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{-37}{45}\)
\(\Rightarrow x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}=-\frac{4}{5}\)
\(\Rightarrow x=\frac{-3}{5}\)
b) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\)
\(\Rightarrow2A=1-\frac{1}{2005}\)
\(\Rightarrow2A=\frac{2004}{2005}\)
\(\Rightarrow A=\frac{1002}{2005}\)
Tính tổng:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
= \(\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003+2005}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{2003}-\frac{1}{2005}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{2005}\right)\)
= \(\frac{1}{2}\cdot\frac{2004}{2005}\)
= \(\frac{1002}{2005}\)
k nha