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 \(x=-329\)

a) 

( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0

\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)

P/s: đợi xíu làm câu b

b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{-1}{x+3}=\frac{1}{2015}\)

\(\Leftrightarrow x+3=-2015\)

\(\Leftrightarrow x=-2018\)

Vậy,.........

A/ Ta có số nào nhân với 0 cx = 0 

Vậy từ đó suy ra 2 trường hợp 

TH1\(4x-9=0\)

\(=>x=\frac{9}{4}\)

TH2 \(2,5+-\frac{7}{3}x=0\)

 \(=>x=\frac{15}{14}\)

Bài làm:

Ta có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=2^x\)

\(\Leftrightarrow\frac{1.2.3.....30.31}{2.2.2.3.2.4.....2.31.2.32}=2^x\)

\(\Leftrightarrow\frac{1}{2^{31}.2^5}=2^x\)

\(\Leftrightarrow\frac{1}{2^{36}}=2^x\)

\(\Rightarrow x=-36\)

mk cần cả giải thích

giúp mk vs!!!

( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0

 <=> +) 1/7 . x - 2/7 = 0                                    +)    (- 1 / 5) . x +3/5 = 0                              +)  1/ 3 . x + 4/ 3 = 0

                    x = 2                                                                  x = 3                                                         x = 4

                                                    Vậy x = 2 : x = 3 ; x=4

\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)

=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)

=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)

=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)=  \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)

=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

=> \(x=10\)

b) Tương tự câu a

Thiếu điều kiện xy = 1; x+y khác 0 nhá bn

Bài này tương tự câu 1 ở đây

Vi \(\left(\frac{1}{7}x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{7}x-\frac{2}{7}=0\\-\frac{1}{5}x+\frac{3}{5}=0\\\frac{1}{3}x+\frac{4}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}\frac{1}{7}x=\frac{2}{7}\\-\frac{1}{5}x=-\frac{3}{5}\\\frac{1}{3}x=-\frac{4}{3}\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=3\\x=-4\end{cases}}}\)

Vậy \(x\in\left\{-4;3;2\right\}\)

\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\text{ hoặc }-\frac{1}{5}x+\frac{3}{5}=0\text{ hoặc }\frac{1}{3}x+\frac{4}{3}=0\)

\(\Rightarrow x=2\text{ hoặc }x=3\text{ hoặc }x=-4\)

Vậy tập nghiệm của pt là \(S=\left\{2;3;-4\right\}\) 

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