Question

Tìm x biết :

a) \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{50}\)

b) \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)

c) \(\frac{x-10}{20}+\frac{x-20}{10}+\frac{x-30}{5}=\frac{x-14}{4}\)

Answer 1

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\) 

\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)

Nên x + 1 = 0

=> x = -1

Answer 2

còn b vs c thì sao ạ

Answer 3

b) Sai đề à bạn đề \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)  hả đề này mk làm đc 

Answer 4

ukm để mik xem lại câu b,vậy bạn làm đc câu c ko

Answer 5

c) \(\frac{x-10}{20}+\frac{x-20}{10}+\frac{x-30}{5}=\frac{x-14}{4}\)

=> \(\frac{x-10}{20}+\frac{2x-40}{20}+\frac{4x-120}{20}=\frac{5x-70}{20}\)

=> \(\frac{x-10}{20}+\frac{2x-40}{20}+\frac{4x-120}{20}-\frac{5x-70}{20}=0\)

=> \(\left(x-10\right)+\left(2x-40\right)+\left(4x-120\right)-\left(5x-70\right)=0\)

=> \(x-10+2x-40+4x-120-5x+70=0\)

=> \(\left(x+2x+4x-5x\right)+\left(70-10-40-120\right)=0\)

=> \(2x+\left(-100\right)=0\)

=> \(2x=100\)

=> \(x=50\)