Bài làm:

c) \(\left(x-2\right)\left(x+3\right)>0\)

Ta xét 2 trường hợp sau:

+ Nếu \(\hept{\begin{cases}x-2>0\\x+3>0\end{cases}\Rightarrow}\hept{\begin{cases}x>2\\x>-3\end{cases}\Rightarrow}x>2\)

+ Nếu \(\hept{\begin{cases}x-2< 0\\x+3< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x< -3\end{cases}}\Rightarrow x< -3\)

Vậy \(\orbr{\begin{cases}x>2\\x< -3\end{cases}}\)

d) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Leftrightarrow3x=\frac{1}{9}\)

\(\Leftrightarrow x=\frac{1}{27}\)

Vậy \(x=\frac{1}{27}\)

Học tốt!!!!

\(1+\frac{-1}{60}+\frac{19}{120}< \frac{x}{36}< \frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)

=> \(\frac{137}{120}< \frac{x}{36}< \frac{521}{360}\)

=> \(\frac{411}{360}< \frac{10x}{360}< \frac{521}{360}\)

=> 411 < 10x < 521

=> x \(\in\){ 42,43,44,...,52}

c) \(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)+\frac{5}{9}=\frac{23}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{5}{9}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{23}{27}-\frac{15}{27}\)

\(\frac{7}{9}\div\left(2+\frac{3}{4}x\right)=\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}\div\frac{8}{27}\)

\(2+\frac{3}{4}x=\frac{7}{9}.\frac{27}{8}\)

\(2+\frac{3}{4}x=\frac{21}{8}\)

\(\frac{3}{4}x=\frac{21}{8}-2\)

\(\frac{3}{4}x=\frac{21}{8}-\frac{16}{8}\)

\(\frac{3}{4}x=\frac{5}{8}\)

\(x=\frac{5}{8}\div\frac{3}{4}\)

\(x=\frac{5}{8}.\frac{4}{3}\)

\(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}\).

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

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

\(\left|x-\frac{1}{3}\right|=\frac{20}{12}+\frac{9}{12}\)

\(\left|x-\frac{1}{3}\right|=\frac{29}{12}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{29}{12}\\x-\frac{1}{3}=-\frac{29}{12}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{11}{4}\\x=-\frac{25}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{11}{4};-\frac{25}{12}\right\}\).

khong can lam nua dau minh biet lam roi

+ Tìm x

\(\frac{3}{x}\)= \(\frac{-36}{84}\)=> x . ( - 36 ) = 3 . 84

                            => x = \(\frac{3.84}{-36}\)=> x = - 7

+ Tìm y

\(\frac{y}{35}\)= \(\frac{-36}{84}\)=> y . 84 = ( - 36 ) . 35

                               => y = \(\frac{\left(-36\right).35}{84}\)=> y = - 15

\(\frac{x}{2}-2=\frac{y}{3}-2=\frac{z}{4}-2\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{27}{9}=3\)
\(\Rightarrow x=6,y=9,z=12\)

Áp dụng tính chất dãy tỉ số bằng nhau ,ta có:

\(\frac{x-4}{2}=\frac{y-6}{3}=\frac{z-8}{4}=\frac{x+y+z-18}{2+3+4}=1\)

Ta có:\(\frac{x-4}{2}=1\Rightarrow x=6\)

\(\frac{y-6}{3}=1\Rightarrow y=9\)

\(\frac{z-8}{4}=1\Rightarrow z=12\)

Áp dụng tính chất tỉ lệ thức, ta có:

\(\frac{x-4}{2}=\frac{y-6}{3}=\frac{z-8}{4}=\frac{x+y+z-18}{9}=\frac{27-18}{9}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{x-4}{2}=1\Leftrightarrow x=6\\\frac{y-6}{3}=1\Leftrightarrow y=9\\\frac{z-8}{4}=1\Leftrightarrow z=12\end{cases}}\)

làm luông di

đây nhá , chị nêu phương pháp cho 

\(\frac{A}{B}\)= \(\frac{C}{D}\) 

<=> A*D=B*C