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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 )

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) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)

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. \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{22}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{22}{45}\)

\(\Rightarrow x=2\)

23 tháng 7 2017

\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{9}{45}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\Rightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)

\(\Rightarrow x=\frac{7}{\frac{21}{45}}=15\)

Vậy \(x=15\).

25 tháng 6 2019

\(a,\frac{x-1}{21}=\frac{3}{x+1}\)

\(\Leftrightarrow\left[x-1\right]\left[x+1\right]=63\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

\(\Leftrightarrow x^2=8^2\)

\(\Leftrightarrow x=\pm8\)

25 tháng 6 2019

\(b,\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left[\frac{1}{5}-\frac{1}{45}\right]=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{21}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{7}{15}\)

\(\Leftrightarrow x=15\)

Vậy x = 15

Bài cuối tương tự

10 tháng 8 2018

\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}\)

\(\Leftrightarrow x=\frac{7.45}{21}=15\)

7 tháng 6 2020

\(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{9}{45}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)

\(\frac{7}{x}=\frac{21}{45}\)

\(\frac{7}{x}=\frac{7}{15}\)

\(\Rightarrow x=15\)

Vậy \(x=15\).

7 tháng 6 2020

\(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)

=> \(\frac{7}{x}+4\left(\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+....+\frac{1}{41\cdot45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+4\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+4\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+4\cdot\frac{32}{45}=\frac{29}{45}\)

=> \(\frac{7}{x}+\frac{128}{45}=\frac{29}{45}\)

=> \(\frac{7}{x}=-\frac{11}{5}\)

Đến đây tự giải quyết :))

23 tháng 3 2018

theo đề bài ta có:

\(x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)

\(x+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{-37}{45}\)

\(x+\frac{8}{45}=\frac{-37}{45}\)

\(x=\frac{-37}{45}-\frac{8}{45}\)

\(x=\frac{-45}{45}=1\)

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

19 tháng 3 2017

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=-1

vậy x=-1