Bài làm :

\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)

\(b\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=0+\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)

\(c\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1\div x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-0\\1\div x=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)

\(d\text{)}...\Leftrightarrow\left(x+3\right)\left(x-4+2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

B = \(\frac{-1}{9}\). \(\frac{-17}{29}\). \(\frac{58}{51}\)= \(\frac{2}{27}\)

D = \(\frac{-184}{9}\)

Nếu cần cách làm cụ thể thì ib tớ nhé, xl vì tớ không bt bấm cách làm cụ thể !!!:))

#Cụ_Maiz

\(B=\frac{-1}{9}.\frac{-17}{29}.\frac{58}{51}\)

\(=\frac{1.27.58}{9.29.51}\)

\(=\frac{1.27.29.2}{9.29.17.3}\)

\(=\frac{2}{27}\)

\(D=\left(1\frac{1}{27}.\frac{12}{23}.\frac{9}{14}\right):\left(\frac{-3}{23}\right)\)

\(=\frac{28}{27}.\frac{12}{23}.\frac{9}{14}.\left(\frac{-23}{3}\right)\)

\(=\frac{7.4.2.2.3.9.\left(-23\right)}{9.3.23.7.2.3}\)

\(=\frac{-8}{3}\)

Học tốt nhé

\(\left(\frac{3}{4}x-\frac{9}{16}\right)\left(1,5+\frac{3}{5}:x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{9}{16}=0\\\frac{3}{2}+\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{4}x=\frac{9}{16}\\\frac{3}{5x}=-\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{2}{5}\end{cases}}\)

 Có 2 trường hợp :

 Trưởng hợp 1 :( \(\frac{3}{4}\)x - \(\frac{9}{16}\)) = 0 

Ta có :  \(\frac{3}{4}\)x  = 0 + \(\frac{9}{16}\)

             \(\frac{3}{4}\)x = \(\frac{9}{16}\)

                      x = \(\frac{9}{16}\): \(\frac{3}{4}\)

               =>  x = \(\frac{3}{4}\)

Trường hợp 2 : ( 1, 5 + \(\frac{3}{5}\): x ) = 0 

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

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

                                          \(\frac{3}{5}\): x = \(\frac{-3}{2}\)

                                                     x = \(\frac{3}{5}\): \(\frac{-3}{2}\)

                                            = x = \(\frac{-2}{5}\)

Vậy : x = \(\frac{3}{4}\)Hoặc x = \(\frac{-2}{5}\)

Study well ^_^:))

# Cụ Maiz 

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)

\(\Leftrightarrow2x+3=93\)

\(\Leftrightarrow2x=90\)

\(\Leftrightarrow x=45\)

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Rightarrow\frac{1}{2x+3}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(\Rightarrow2x=90\)

\(\Rightarrow x=45\)

Vậy x = 45.

a) \(\frac{x-1}{21}=\frac{3}{x+1}\)( ĐKXĐ : x khác -1 )

<=> ( x - 1 )( x + 1 ) = 21.3

<=> x² - 1 = 63

<=> x² = 64

<=> x² = ( ±8 )²

<=> x = ±8 ( tmđk )

b) \(\frac{7}{x}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}=\frac{29}{45}\)( ĐKXĐ : x khác 0 )

<=> \(\frac{7}{x}+\left(\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}\right)=\frac{29}{45}\)

<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

<=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

<=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

<=> \(\frac{7}{x}=\frac{7}{15}\)

<=> x = 15 ( tmđk )

a) \(\frac{x-1}{21}=\frac{3}{x+1}\Leftrightarrow\left(x-1\right)\left(x+1\right)=3.21\)

\(\Leftrightarrow x^2-1=63\Rightarrow x^2=63+1=64\Rightarrow x=\pm8\)

b) \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\Leftrightarrow\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\Leftrightarrow\frac{7}{x}=\frac{29}{45}-\frac{8}{45}=\frac{21}{45}=\frac{7}{15}\Rightarrow x=15\)

            Bài làm :

Ta có :

\(\left|x+\frac{4}{15}\right|-\left|3,75\right|=-\left|2,15\right|\)

\(\Leftrightarrow\left|x+\frac{4}{15}\right|-3,75=-2,15\)

\(\Leftrightarrow\left|x+\frac{4}{15}\right|=1,6\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{4}{15}=1,6\\x+\frac{4}{15}=-1,6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{28}{15}\end{cases}}\)

Vậy x=4/3 hoặc x=-28/15

cảm ơn ~~

4.5^7= 4.78125=312500

(vì 5^7= 5.5.5.5.5.5.5)

\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right)\)

\(=\frac{3}{5}+\frac{3}{11}+\frac{3}{7}+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}-\frac{23}{44}\)

\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)

\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)

\(=\frac{35}{35}+\left(\frac{-44}{44}\right)+\frac{2}{97}=1+\left(-1\right)+\frac{2}{97}=\frac{2}{97}\)

\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\) 

\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\) 

\(=-1+1+\frac{2}{97}\) 

\(=\frac{2}{97}\) 

\(=\left(-\frac{1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{2}{7}+\frac{4}{35}\right)+\frac{1}{127}\) 

\(=\left(-\frac{9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{10}{35}+\frac{4}{35}\right)+\frac{1}{127}\) 

\(=\left(-\frac{18}{18}\right)+\frac{35}{35}+\frac{1}{127}\) 

\(=-1+1+\frac{1}{127}\) 

\(=\frac{1}{127}\)

\(\frac{1}{5}< x< 1\frac{1}{5}\Rightarrow\hept{\begin{cases}x-\frac{1}{5}>0\\1\frac{1}{5}-x>0\end{cases}\Rightarrow\hept{\begin{cases}\left|x-\frac{1}{5}\right|=x-\frac{1}{5}\\\left|1\frac{1}{5}-x\right|=1\frac{1}{5}-x\end{cases}}}\)

\(\Rightarrow M=1\frac{1}{5}-x+x-\frac{1}{5}-3\frac{1}{5}=1\frac{1}{5}-\frac{1}{5}-3\frac{1}{5}=\frac{-11}{5}\)