a) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}:x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}:x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)

Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)

b) 

 \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

        \(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}+4=4\)

         \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

          \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)

\(\Rightarrow x+329=0\)

\(\Rightarrow x=-329\)

Vậy \(x=-329\)

d, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow x+10=0\) (Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\) ≠ 0)

\(\Leftrightarrow x=-10\)

Vậy x = -10 là nghiệm của phương trình.

e, Đề sai.

Sửa đề: \(\frac{x}{2016}+\frac{x+1}{2017}+\frac{x+2}{2018}+\frac{x+3}{2019}+\frac{x+4}{2020}=5\)

\(\Leftrightarrow\frac{x}{2016}-1+\frac{x+1}{2017}-1+\frac{x+2}{2018}-1+\frac{x+3}{2019}-1+\frac{x+4}{2020}-1=0\)

\(\Leftrightarrow\frac{x-2016}{2016}+\frac{x-2016}{2017}+\frac{x-2016}{2018}+\frac{x-2016}{2019}+\frac{x-2016}{2020}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\right)=0\)

\(\Leftrightarrow x-2016=0\) (Vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\ne0\)

\(\Leftrightarrow x=2016\)

Vậy x = 2016 là nghiệm của phương trình.

Giúp mình với các bạn ơi!!!!!!!!!!!!!!!!!!!!!!!!!!!!

a) \(\frac{x-1}{6}=\frac{2x+3}{7}\)

\(\Leftrightarrow7\left(x-1\right)=6\left(2x+3\right)\)

\(\Leftrightarrow7x-7=12x+18\)

\(\Leftrightarrow5x+18=-7\)

\(\Leftrightarrow5x=-25\)

\(\Leftrightarrow x=-5\)

b) \(\left(2x^2-\frac{1}{2}x\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{2}\right)\left(x^2+1\right)=0\)

Vì \(x^2+1>0\)nên \(\orbr{\begin{cases}x=0\\2x-\frac{1}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

c) \(\frac{4}{3}-\frac{1}{3}\left|x-1\right|=\frac{1}{2}\)

=> \(\frac{1}{3}\left|x-1\right|=\frac{4}{3}-\frac{1}{2}\)

=> \(\frac{1}{3}\left|x-1\right|=\frac{5}{6}\)

=> \(\left|x-1\right|=\frac{5}{6}:\frac{1}{3}\)

=> \(\left|x-1\right|=\frac{5}{2}\)

=> \(\orbr{\begin{cases}x-1=\frac{5}{2}\\x-1=-\frac{5}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{3}{2}\end{cases}}\)

\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}x=0\)

\(x.\left(\frac{1}{3}+\frac{2}{5}\right)-\frac{2}{5}=0\)

\(x.\frac{11}{15}=0+\frac{2}{5}=\frac{2}{5}\)

\(x=\frac{2}{5}:\frac{11}{15}=\frac{6}{11}\)

Mình không biết! Mình mới lớp 5 thôi!

=1/3*x+2/5*x-2/5.

=x*(1/3+2/5)-2/5.

=x*11/15-2/5=0.

=>x*11/15=2/5

=>x=2/5:11/15=6/11.

k nha có j kb vs mk