Question

Tìm x :

a. x : \(\frac{3}{4}\) = 2x \(\times\) \(\frac{7}{3}\) + 5

b. 3x - 2 = 5x - 4

c. ( \(\frac{1}{1\times2}\)+ \(\frac{1}{2\times3}\)+ \(\frac{1}{3\times4}\)+ .... + \(\frac{1}{11\times12}\) ) \(\times\) x = 2

d. x : ( \(\frac{1}{1\times3}\) + \(\frac{1}{3\times5}\) + ..... + \(\frac{1}{101\times103}\) ) = 1

 

Answer 1

a, \(\frac{4x}{3}=\frac{14x}{3}+5\)

\(\frac{4x}{3}-\frac{14x}{3}=5\) 

\(\frac{-10x}{3}=5\)

x=-1,5

b, 3x-5x=2-4

-2x=-2

x=1

c, \(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\right)\)  .x=2

\(\left(\frac{1}{1}-\frac{1}{12}\right).x=2\)

\(\frac{11}{12}.x=2\)

x=\(\frac{24}{11}\)

d, \(x=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{101.103}\right)\)

\(x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{103}\right)\)

x=\(\frac{1}{2}\left(1-\frac{1}{103}\right)\)

x=\(\frac{1}{2}.\frac{102}{103}\)

x=\(\frac{51}{103}\)