câu 2

\(\dfrac{2a}{1}\)=\(\dfrac{3b}{1}\)=\(\dfrac{4c}{1}\)=\(\dfrac{2a}{12}\)=\(\dfrac{3b}{12}\)=\(\dfrac{4c}{12}\)

\(\Rightarrow\)\(\dfrac{a}{6}\)=\(\dfrac{b}{4}\)=\(\dfrac{c}{3}\)

Theo tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{6}\)=\(\dfrac{b}{4}\)=\(\dfrac{c}{3}\)=\(\dfrac{a+b-c}{6+4-3}\)=\(\dfrac{14}{7}\)=2

\(\)\(\dfrac{a}{6}\)=\(\dfrac{2}{1}\)thì a=2.6:1=12

\(\dfrac{b}{4}\)=\(\dfrac{2}{1}\)thì b=2.4:1=8

\(\dfrac{c}{3}\)=\(\dfrac{2}{1}\)thì c=2.3:1=6

\(\dfrac{25}{26}\)=\(\dfrac{25\cdot10100}{26\cdot10100}\)=\(\dfrac{25250}{26260}\)

Ta có 1-\(\dfrac{25250}{26260}\)=\(\dfrac{1010}{26260}\);1-\(\dfrac{25251}{26261}\)=\(\dfrac{1010}{26261}\)

Vì 26260<26261 nên \(\dfrac{1010}{26260}\)>\(\dfrac{1010}{26261}\)

Vậy \(\dfrac{25}{26}\)>\(\dfrac{25251}{26261}\)

\(\dfrac{1}{3}\)x(\(\dfrac{3}{1+4}\)+\(\dfrac{3}{4+7}\)+........+\(\dfrac{3}{101+103}\))

\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{7}\)+.........+\(\dfrac{ }{ }\)\(\dfrac{1}{101}\)-\(\dfrac{1}{103}\))

\(\dfrac{1}{3}\)x(\(\dfrac{1}{1}\)-\(\dfrac{1}{103}\))

\(\dfrac{1}{3}\)x\(\dfrac{102}{103}\)

\(\dfrac{34}{103}\)

3/8+4/1+3/8=(3/8+3/8)+4/1=3/4+4/1

=3/4+12/4=15/4

yes

y= 6,2,-2,-10