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NP
7
23 tháng 3 2018
theo đề bài ta có:
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)
\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=1\)
TC
23 tháng 3 2018
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
19 tháng 3 2017
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
25 tháng 3 2017
\(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=-\dfrac{37}{45}\\ x+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{44}-\dfrac{1}{45}\right)=-\dfrac{37}{45}\\ x+\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=-\dfrac{37}{45}\\ x+\dfrac{8}{45}=-\dfrac{37}{45}\\ x=-\dfrac{37}{45}-\dfrac{8}{45}\\ x=-1\)
QD
6 tháng 6 2019
\(x+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(\Leftrightarrow x+3\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}.\frac{8}{45}=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{2}{15}=-\frac{37}{45}\)
\(\Leftrightarrow x=-\frac{43}{45}\)
NN
0
DN
0
LP
7 tháng 4 2015
Theo đề bài ta có
x+1/5-1/9+1/9-1/13+.........+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
x=-1
NL
14 tháng 4 2017
Theo bài ra ta có:
x1/5-1/9+1/9-1/13+...+1/41-1/45=-37/45
x+8/45=-37/45
x=-45/45
x=-1/1
x=-1
M
1
DA
5
B
23 tháng 3 2017
\(x+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{-37}{45}\)
\(x+4\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{4}{13.17}+...+\frac{1}{41.45}\right)=\frac{-37}{45}\)
\(x+4.\frac{1}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(x+1\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(x+\frac{8}{45}=\frac{-37}{45}\)
\(x=\frac{-37}{45}-\frac{8}{45}\)
\(x=\frac{-45}{45}=-1\)
\(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=-\dfrac{37}{45}\)
=>\(x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=-\dfrac{37}{45}\)
=>\(x+\dfrac{1}{5}-\dfrac{1}{45}=-\dfrac{37}{45}\)
=>\(x+\dfrac{8}{45}=-\dfrac{37}{45}\)
=>\(x=-\dfrac{37}{45}-\dfrac{8}{45}=-\dfrac{45}{45}=-1\)
\(x+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\cdots+\frac{4}{41\cdot45}=-\frac{37}{45}\)
\(\Rightarrow4x+\frac{4\cdot4}{5\cdot9}+\frac{4\cdot4}{9\cdot13}+\cdots+\frac{4\cdot4}{41\cdot45}=\frac{\left(-37\right)\cdot4}{45}\) (1)
Ta có: \(\frac45-\frac49=\frac{36-20}{45}=\frac{16}{45}=\frac{4\cdot4}{5\cdot9}\)
Tương tự: \(\frac49-\frac{4}{13}=\frac{4\cdot13-4\cdot9}{9\cdot13}=\frac{4\cdot4}{9\cdot13}\)
...
Khi đó, (1) trở thành:
\(4x+\frac45-\frac49+\frac49-\frac{4}{13}+\cdots+\frac{4}{41}-\frac{4}{45}=-\frac{148}{45}\)
\(4x+\frac45-\left(\frac49-\frac49\right)-\left(\frac{4}{13}-\frac{4}{13}\right)-\cdots-\frac{4}{45}=-\frac{148}{45}\)
\(4x+\frac45-\frac{4}{45}=-\frac{148}{45}\)
\(4x+\frac{4\cdot9-4}{45}=-\frac{148}{45}\)
\(4x+\frac{32}{45}=-\frac{148}{45}\)
\(4x=-\frac{148}{45}-\frac{32}{45}\)
\(4x=-\frac{180}{45}=-4\)
\(x=-1\)
Vậy \(x=-1\)