Tìm x

 
$$$a)4\frac{1}{3}(\frac{1}{2}-\frac{1}{6})\le x\le \frac{-2}{3}(\frac{1}{3}-\frac{1}{2}-\frac{3}{4})=x+\frac{1}{5}$​​​​​​​
$b)\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}$
$$\begin{gathered} c)x-128=\left(40\frac{21}{20}-5\right)÷\left(\frac{4141}{4242}-1\right)÷\left(\frac{636363}{646464}=1\right) \\ d)4x(x-5)-x(4x-7)=26 \end{gathered}$$
$e)\left(2x-3\right)\left(3x+1\right)=\left(\frac{8}{2017}-\frac{8}{2018}+27\frac{3}{2019}\right)\left(-\frac{34}{35}+\frac{2}{5}-\frac{3}{7}+1\right)$
$g)\left(\frac{14098}{13.20}+\frac{14098}{20.27}+...+\frac{14098}{62.69}\right)x=\frac{2013}{2}\left(\frac{-5}{9.13}-\frac{7}{9.25}-\frac{13}{19.25}-\frac{31}{19.69}\right)$
Hết rùi mong các bạn làm nhanh hộ mình nha