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- B = \(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{41.45}.\)
K
Khách
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Những câu hỏi liên quan
NH
4
26 tháng 4 2017
\(x+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=1\)
\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=1\)
\(x+\frac{1}{5}-\frac{1}{45}=1\)
\(x+\frac{8}{45}=1\)
\(\Rightarrow x=1-\frac{8}{45}\)
\(\Rightarrow x=\frac{37}{45}\)
Q
26 tháng 4 2017
\(x+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=1\)
\(x+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=1\)
\(x+\left[4\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{41.45}\right)\right]=1\)
\(x+\left[4.\frac{1}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\right]=1\)
\(x+\left[1\left(\frac{1}{5}-\frac{1}{45}\right)\right]=1\)
\(x+\frac{8}{45}=1\)
\(x=1-\frac{8}{45}\)
\(x=\frac{37}{45}\)
17 tháng 9 2020
a) \(\frac{x-1}{21}=\frac{3}{x+1}\)( ĐKXĐ : x khác -1 )
<=> ( x - 1 )( x + 1 ) = 21.3
<=> x2 - 1 = 63
<=> x2 = 64
<=> x2 = ( ±8 )2
<=> x = ±8 ( tmđk )
b) \(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)( ĐKXĐ : x khác 0 )
<=> \(\frac{7}{x}+\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
<=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
<=> \(\frac{7}{x}=\frac{7}{15}\)
<=> x = 15 ( tmđk )
VC
Vũ Cao Minh⁀ᶦᵈᵒᶫ ( ✎﹏IDΣΛ亗 )
CTV
VIP
17 tháng 9 2020
a) \(\frac{x-1}{21}=\frac{3}{x+1}\Leftrightarrow\left(x-1\right)\left(x+1\right)=3.21\)
\(\Leftrightarrow x^2-1=63\Rightarrow x^2=63+1=64\Rightarrow x=\pm8\)
b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)
KN
9 tháng 10 2020
e. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)
\(\Rightarrow\frac{7}{x}=\frac{7}{15}\)
\(\Rightarrow x=15\)
f. \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)
\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{22}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{22}{45}\)
\(\Rightarrow x=2\)
NT
4
28 tháng 6 2017
Ta có : \(x+\frac{1}{1.5}+x+\frac{1}{5.9}+x+\frac{1}{9.13}+......+x+\frac{1}{397.401}=101x\)
\(\Leftrightarrow\left(x+x+x+......+x\right)+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)
\(\Leftrightarrow100x+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)
\(\Rightarrow x=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\)
\(\Rightarrow4x=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{397.401}\)
\(\Rightarrow4x=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.....+\frac{1}{397}-\frac{1}{401}\)
\(\Rightarrow4x=1-\frac{1}{401}\)
\(\Rightarrow4x=\frac{400}{401}\)
\(\Rightarrow x=\frac{400}{401}.\frac{1}{4}=\frac{100}{401}\)
28 tháng 6 2017
tui biết giải, mà k biết có bao nhiêu x, bạn tính sao ra 100x vậy bạn?
20 tháng 3 2022
\(A=3\times\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{101}-\frac{1}{105}\right)\)
\(A=3\times\left(1-\frac{1}{105}\right)\)
\(A=3\times\frac{104}{105}\)
\(A=\frac{104}{35}\)
MA
1
QD
6 tháng 6 2019
\(x+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+...+\frac{4}{41.45}=-\frac{37}{45}\)
\(\Leftrightarrow x+3\left(\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}\left(\frac{1}{5}-\frac{1}{45}\right)=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{3}{4}.\frac{8}{45}=-\frac{37}{45}\)
\(\Leftrightarrow x+\frac{2}{15}=-\frac{37}{45}\)
\(\Leftrightarrow x=-\frac{43}{45}\)
6 tháng 3 2016
A=(1/2-1/5+1/5-1/8+1/8-....+1/50-1/51)
= 1/2-1/51
= 51/102 - 2/102
= 49/102
B=1.4/1.5.4+1.4/5.9.4+...+1.4/41.45.4
= 1/4(1-1/5+1/5-1/9+1/9-...+1/41-1/45)
= 1/4(1-1/45)
= 1/4.44/45
= 11/45
S
10 tháng 5 2017
\(A=8400\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(=\frac{8400}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}+\frac{4}{21.25}\right)\)
\(=2100\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)
\(=2100\left(1-\frac{1}{25}\right)\)
\(=2100\cdot\frac{24}{25}\)
\(=2016\)
10 tháng 5 2017
\(A=8400.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\left(\frac{1.4}{1.5.4}+\frac{1.4}{5.9.4}+\frac{1.4}{9.13.4}+\frac{1.4}{13.17.4}+\frac{1.4}{17.21.4}+\frac{1.4}{21.25.4}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+\frac{1}{17.21}+\frac{1}{21.25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}+\frac{1}{21}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{25}\right)\)
\(A=8400.\frac{1}{4}.\frac{24}{25}\)
\(A=2016\)
\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(4B=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\)
\(4B=\frac{44}{45}\)
\(B=\frac{11}{45}\)
\(B=\frac{1}{1.5}+\frac{1}{5.9}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{41.45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)