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

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

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

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

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

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

\(\Leftrightarrow x^2=64\)

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

\(\Leftrightarrow x=\pm8\)

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

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

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

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

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

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

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

\(\Leftrightarrow x=15\)

Vậy x = 15

Bài cuối tương tự

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

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

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

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

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

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

Vậy \(x=15\).

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow x=15\)

Vậy \(x=15\).

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

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

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

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

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

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

=> \(\frac{7}{x}=-\frac{11}{5}\)

Đến đây tự giải quyết :))

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

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

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

\(\Leftrightarrow\frac{1}{2}x=1\)

\(\Rightarrow x=2\)

=>(2/1.2.3+2/2.3.4+....+2/8.9.10).x=22/45

=>(1/1.2-1/2.3+1/2.3-1/3.4+....+1/8.9-1/9.10).x=22/45

=>(1/1.2-1/9.10).x=22/45

=>22/45.x=44/45

=>x=2

4+2^2+2^3+....+2^20=2^n

=>2^2+2^2+2^3+....+2^20=2^n

đặt 2^2+2^3+....+2^20

=>2A-A=2^21-2^2

khi đó A=2^2+2^21-2^2=2^21=2^n

=>n=21

Đặt B = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{8.9.10}\)

=> 2B = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{8.9.10}\)

=> 2B = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\)

=> 2B = \(\frac{1}{1.2}-\frac{1}{9.10}\)

2B = \(\frac{22}{45}\)

B = \(\frac{22}{45}:2\)

=> B = \(\frac{11}{45}\)

Ta có : \(\frac{11}{45}.x=\frac{22}{45}\)

=> x = \(\frac{22}{45}:\frac{11}{45}\)

=> x = \(\frac{2}{1}\)