\(\frac{x+1}{49}+1+\frac{x+2}{48}+1+\frac{x+3}{47}+1+\frac{x+4}{46}+1+\frac{x+5}{45}+1=0\)

\(\Leftrightarrow\frac{x+50}{49}+\frac{x+50}{48}+...+\frac{x+50}{45}=0\)

\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{45}\right)=0\)

Vì 1/49+1/48+...+1/45 khác 0

Nên x+50=0

do đó x=-50

Ta có : 

\(\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-4}{46}+\frac{x-5}{45}=5\)

\(\Leftrightarrow\)\(\left(\frac{x-1}{49}-1\right)+\left(\frac{x-2}{48}-1\right)+\left(\frac{x-3}{47}-1\right)+\left(\frac{x-4}{46}-1\right)+\left(\frac{x-5}{45}-1\right)=5-5\)

\(\Leftrightarrow\)\(\frac{x-1-49}{49}+\frac{x-2-48}{48}+\frac{x-3-47}{47}+\frac{x-4-46}{46}+\frac{x-5-45}{45}=0\)

\(\Leftrightarrow\)\(\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{46}+\frac{x-50}{45}=0\)

\(\Leftrightarrow\)\(\left(x-50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)

Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\) ( vì nó lớn hơn 0 ) 

Nên \(x-50=0\)

\(\Rightarrow\)\(x=50\)

Vậy \(x=50\)

Chúc bạn học tốt ~ 

cảm ơn bạn Phùng Minh Quân

c) \(\dfrac{7x-1}{2}=5+\dfrac{9-5x}{6}\)

\(\Leftrightarrow\dfrac{6\left(7x-1\right)}{12}=\dfrac{5\cdot12}{12}+\dfrac{2\left(9-5x\right)}{12}\)

\(\Rightarrow42x-6=60+18-10x\)

\(\Leftrightarrow52x-84=0\)

\(\Leftrightarrow x=\dfrac{21}{13}\)

Vậy....

d) tương tự

a) \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)ĐKXĐ : \(x\ne2;4\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)

\(\Leftrightarrow\dfrac{2x^2-11x+16}{x^2-6x+8}=-1\)

\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Leftrightarrow3x^2-9x-8x+24=0\)

\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)( thỏa mãn ĐKXĐ )

Vậy....

b) \(\dfrac{x+16}{49}+\dfrac{x+18}{47}=\dfrac{x+20}{45}-1\)

\(\Leftrightarrow\dfrac{x+16}{49}+1+\dfrac{x+18}{47}+1=\dfrac{x+20}{45}-1+1+1\)

\(\Leftrightarrow\dfrac{x+16+49}{49}+\dfrac{x+18+47}{47}=\dfrac{x+20+45}{45}\)

\(\Leftrightarrow\dfrac{x+65}{49}+\dfrac{x+65}{47}=\dfrac{x+65}{45}\)

\(\Leftrightarrow\left(x+65\right)\left(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\right)=0\)

Vì \(\dfrac{1}{49}+\dfrac{1}{47}-\dfrac{1}{45}\ne0\)

\(\Leftrightarrow x+65=0\)

\(\Leftrightarrow x=-65\)

Vậy....

1, Tính tổng:

\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)

\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)=\frac{5}{7}\cdot\frac{-7}{11}=\frac{-5}{11}\)

2, Tìm x:

\(x+\frac{5}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\Rightarrow x+\frac{9}{45}-\frac{1}{45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\Rightarrow x=\frac{-37}{45}-\frac{8}{45}=\frac{-45}{45}=-1\)

- Các bài tìm x còn lại bạn cứ theo trình tự thực hiện phép tính mà làm nhé!

\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)

\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)

\(=\frac{5}{7}\cdot-\frac{7}{11}\)

\(=-\frac{5}{11}\)

\(C=\frac{5}{7}.\frac{5}{11}+\frac{5}{7}.\frac{2}{11}-\frac{5}{7}.\frac{14}{11}\)

\(\Leftrightarrow C=\frac{5}{7}\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)

\(\Leftrightarrow C=\frac{5}{7}\times\frac{-7}{11}\)

\(\Leftrightarrow C=\frac{-35}{77}=\frac{-5}{11}\)

lại ăn gian tiếp

Giúp mk điPeter Jin

a, <=> (59-x/41 + 1) + (57-x/43 + 1) + (55-x/45 + 1) + (53-x/47 + 1) + (51-x/49 + 1) = 0

<=> 100-x/41 + 100-x/43 + 100-x/45 + 100-x/47 + 100-x/49 = 0

<=> (100-x).(1/41+1/43+1/45+1/47+1/49) = 0

<=> 100-x=0 ( vì 1/41+1/43+1/45+1/47+1/49 > 0 )

<=> x=100

Vậy x = 100

b, <=> 2-x/2016 + 1 = (1-x/2017 + 1) + (1 - x/2018)

<=> 2018-x/2016 = 2018-x/2017 + 2018-x/2018

<=> 2018-x/2016 - 2018-x/2017 - 2018-x/2018 = 0

<=> (2018-x).(1/2016-1/2017-1/2018) = 0

<=> 2018-x=0 ( vì 1/2016-1/2017-1/2018 khác 0 )

<=> x=2018

Vậy x=2018

Tk mk nha