Ta có : \(x+\frac{1}{1.5}+x+\frac{1}{5.9}+x+\frac{1}{9.13}+......+x+\frac{1}{397.401}=101x\)

\(\Leftrightarrow\left(x+x+x+......+x\right)+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)

\(\Leftrightarrow100x+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)

\(\Rightarrow x=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\)

\(\Rightarrow4x=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{397.401}\)

\(\Rightarrow4x=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.....+\frac{1}{397}-\frac{1}{401}\)

\(\Rightarrow4x=1-\frac{1}{401}\)

\(\Rightarrow4x=\frac{400}{401}\)

\(\Rightarrow x=\frac{400}{401}.\frac{1}{4}=\frac{100}{401}\)

tui biết giải, mà k biết có bao nhiêu x, bạn tính sao ra 100x vậy bạn?

Đơn giải thôi bạn chỉ cần lấy công thức tính số số hạng là ra thôi

(397 - 1) : 4 + 1 = 100 (số)

\(4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x.\left(x+4\right)}\)

\(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}\) 

\(4A=1-\frac{1}{x+4}\) 

\(4A=\frac{x+4-1}{x+4}\)   

\(A=\frac{x+3}{\text{4(x+4)}}\)

Bạn tự thay rồi tính nhé 

\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+........+\frac{1}{x\cdot\left(x+4\right)}\)

\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+........+\frac{4}{x\cdot\left(x+4\right)}\)

\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.......+\frac{1}{x}-\frac{1}{x+4}\)

\(4A=1-\frac{1}{x+4}\)

\(A=\left(1-\frac{1}{x+4}\right):4\)

Khi x = 12 => \(A=\left(1-\frac{1}{12+4}\right):4\)

A = \(\left(1-\frac{1}{16}:4\right)\)

A = \(\frac{15}{16}:4=\frac{15}{64}\)

Khi x = 2 => \(A=\left(1-\frac{1}{2+4}\right):4\)

A = \(\left(1-\frac{1}{6}\right):4\)

A \(=\frac{5}{6}:4=\frac{5}{24}\)

Khi x = \(\frac{5}{6}\)=> \(A=\left(1-\frac{1}{\frac{5}{6}+4}\right):4\)

A = \(\left(1-\frac{1}{\frac{29}{6}}\right):4\)

A = \(\frac{23}{29}:4=\frac{23}{116}\)

câu \(\frac{5}{6}\)làm thế nào

Ta có : \(\frac{7}{x-2005}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x-2005}=\frac{29}{45}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)\)

\(\Rightarrow\frac{7}{x-2005}=\frac{29}{45}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)

\(\Rightarrow\frac{7}{x-2005}=\frac{29}{45}-\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}-\frac{8}{45}=\frac{7}{15}\)

\(\Rightarrow x-2005=15\Rightarrow x=15+2005=2020\)

Vậy x =2020

sry =29/45 nha

cái j bằng 29/ 45 cơ

theo đề bài ta có:

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

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

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

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

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

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

x+ 1/5.9 + ... + 1/41+ 1/45 = -37/45

x  + 1/5 - 1/9 + 1/9 -  ......+ 1/41 - 1/45 = -37/45

x+ 1/5 - 1/45 = -37 45

x+ 8/45 =-37/45

x            = -37 /45 - 8/45

x                = -1

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

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

x + 4/5x9 + 4/9x13 + ... + 4/41x45 = -37/45

<=> x + 4/4 . ( 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 = -37/8

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

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

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

\(\frac{x-1}{21}=\frac{3}{\left(x+1\right)}\)

=> (x + 1)(x - 1)  = 3.21

=> x² + x - x - 1 = 63

=> x² - 1 = 63

=> x² = 64

=> x = \(\pm\)8

b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\left(đk:x\ne0\right)\)

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

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

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

=> \(\frac{7}{x}=\frac{7}{15}\)

=> x = 15

Vậy x = 15

\(b)\) \(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{97.101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(1-\frac{1}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(\frac{100}{101}=\frac{2x+4}{101}\)

\(\Leftrightarrow\)\(100=2x+4\)

\(\Leftrightarrow\)\(2x=96\)

\(\Leftrightarrow\)\(48\)

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 ~ 

mình cũng đang bí

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

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

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

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

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

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

Vậy \(x=15\).

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

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

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

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

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

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