a)\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+x:\frac{1}{3}=-4\)

\(\Rightarrow\frac{1}{120}.120+x:\frac{1}{3}=-4\)

\(\Rightarrow1+x:\frac{1}{3}=-4\)

\(\Rightarrow x:\frac{1}{3}=-4-1=-5\)

\(\Rightarrow x=-5.\frac{1}{3}=\frac{-5}{3}\)

b)\(1\frac{3}{5}+\left(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\right).x=\frac{16}{5}\)

\(\Rightarrow\frac{8}{5}+\left[\frac{2.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5.\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\right].x=\frac{16}{5}\)

\(\Rightarrow\frac{8}{5}+\frac{2}{5}.x=\frac{16}{5}\)

\(\Rightarrow\frac{2}{5}.x=\frac{16}{5}-\frac{8}{5}=\frac{8}{5}\)

\(\Rightarrow x=\frac{8}{5}:\frac{2}{5}=\frac{8}{5}.\frac{5}{2}=\frac{8}{2}=4\)

\(\Rightarrow x=4\)

\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)

\(\Leftrightarrow\)\(\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)

\(\Leftrightarrow\)\(\left(\frac{1}{24}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)

\(\Leftrightarrow\)\(1+3y=-4\)

\(\Leftrightarrow\)\(3y=-5\)

\(\Leftrightarrow\)\(y=-\frac{5}{3}\)

Vậy...

Ta có : 

\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y.3=-4\)

\(\Rightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+y.3=-4\)

\(\Rightarrow\left(\frac{5}{120}-\frac{4}{120}\right).120+y.3=-4\)

\(\Rightarrow\frac{1}{120}.120+y.3=-4\)

\(\Rightarrow1+y.3=-4\)

\(\Rightarrow3y=-4-1\)

\(\Rightarrow3y=-5\)

\(\Rightarrow y=-\frac{5}{3}\)

Vậy \(y=-\frac{5}{3}\)

\(\left(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\right).120+y:\frac{1}{3}=-4.\)

\(\Leftrightarrow\left(\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\right).120+y:\frac{1}{3}=-4\)

\(\Leftrightarrow\left(\frac{1}{24}-\frac{1}{30}\right).120+y.\frac{3}{1}=-4\)

\(\Leftrightarrow\frac{1}{120}.120+3y=-4\)

\(\Leftrightarrow1+3y=-4\Leftrightarrow3y=-5\)

\(\Leftrightarrow y=\frac{-5}{3}\)

\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)

\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)

\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3x+2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{2}-\frac{1}{3x+5}=\frac{12}{25}\)

\(\frac{1}{3x+5}=\frac{1}{50}\)

=> 3x+5 = 50

3x = 45

x = 15

y=-5/3 

y=-5/3

y=-5/3

=> 1/24 - 1/25 + 1/25 - 1/ 26 + .... + 1/29 - 1/30 + x : 1/3 = -4

=> 1/24 - 1/30 + x : 1/3 = - 4

=> 1/ 120 + x : 1/3 = -4

=> x : 1/3 = 481/120

=> x = 481/360 

Vậy x = 481/360

\(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}+x:\frac{1}{3}=-4\)

\(\Rightarrow\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}+x\times3=-4\)

\(\Rightarrow\frac{1}{24}-\frac{1}{30}+3x=-4\)

\(\Rightarrow\frac{1}{120}+3x=-4\)

=(1/24-1/25+1/25-1/24+...+1/29-1/30)+x:1/3=-4

=(1/24-1/30)+x:1/3=-4

=1/120+x:1/3=-4

x:1/3=-4-1/120

x:1/3=-481/120

x=-481/120.1/3

x=-481/360

số to lắm bạn nhé!

Giải phương tình nha :v 

a) \(\frac{7x}{8}-5\left(x-9\right)=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{7x}{8}-\frac{40\left(x-9\right)}{8}=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{7x}{8}-\frac{40x-360}{8}=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{360-33x}{8}=\frac{20x+1,5}{6}\)

\(\Leftrightarrow2160-198x=160x+12\)

\(\Leftrightarrow358x=2148\)

\(\Leftrightarrow x=6\)

Vậy nghiệm của pt x=6

b)  \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

\(\Leftrightarrow\frac{10\left(x-1\right)+4}{12}-\frac{21x-3}{12}=\frac{4x+2}{7}-\frac{35}{7}\)

\(\Leftrightarrow\frac{-11x-3}{12}=\frac{4x-33}{7}\)

\(\Leftrightarrow-77x-21=48x-396\)

\(\Leftrightarrow125x=375\)

\(\Leftrightarrow3\)

Vậy nghiệm của pt x=3

c)\(\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)

\(\Leftrightarrow\frac{15\left(x-3\right)}{20}+\frac{8x-21}{20}=\frac{3x+3}{5}+\frac{30}{5}\)

\(\Leftrightarrow\frac{23x-66}{20}=\frac{3x+33}{5}\)

\(\Leftrightarrow115x-330=60x+660\)

\(\Leftrightarrow55x=990\)

\(\Leftrightarrow x=18\)

Vậy nghiệm của pt x=18

d) \(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2\left(3x-1\right)}{5}-\frac{3x+2}{10}\)

\(\Leftrightarrow\frac{6x+3}{4}-\frac{20}{4}=\frac{4\left(3x-1\right)}{10}-\frac{3x+2}{10}\)

\(\Leftrightarrow\frac{6x-17}{4}=\frac{9x-6}{10}\)

\(\Leftrightarrow60x-170=36x-24\)

\(\Leftrightarrow24x=146\)

\(\Leftrightarrow x=\frac{73}{12}\)

Vậy nghiệm của pt \(x=\frac{73}{12}\)