9) \(\dfrac{x}{4}=\dfrac{9}{x}\)

Theo định nghĩa về hai phân số bằng nhau, ta có:

\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

8)

\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)

7)

\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)

6)

\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)

5)

\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)

4)

\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

3)

\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)

2)

\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)

1)

\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)

câu a sai đề nha

a, \(\frac{x}{5}=\frac{2}{3}\Leftrightarrow x=\frac{10}{3}\)

b, \(\frac{x}{-24}=\frac{20}{42}\Leftrightarrow x=-\frac{80}{7}\)

c, \(\frac{x+3}{15}=\frac{1}{3}\Leftrightarrow3x+9=15\Leftrightarrow x=2\)

d, \(\frac{2}{3}x-\frac{3}{2}x=\frac{5}{12}\Leftrightarrow x\left(\frac{2}{3}-\frac{3}{2}\right)=\frac{5}{12}\Leftrightarrow-\frac{5}{6}x=\frac{5}{12}\Leftrightarrow x=-\frac{1}{2}\)

\(2^{x+4}+2^{x+3}+2^{x+2}+2^{x+1}=1920\)

\(15.2^{x+1}=1920\)

\(2^{x+1}=1920:15\)

\(2^{x+1}=128\)

\(2^{x+1}=2^7\)

\(x+1=7\)

\(x=7-1\)

\(x=6\)

=> x = 6

a/ \(\frac{x+2}{27}=\frac{x}{9}\)

=> 9(x + 2) = 27x

=> 9x + 18 = 27x

=> 9x + 18 - 27x = 0

=> 9x - 27x + 18 = 0

=> -18x = -18

=> x = 1

b/ \(\frac{-7}{x}=\frac{21}{34-x}\)

=> -7(34 - x) = 21x

=>  -238 + 7x = 21x

=> 21x - 7x = -238

=> -14x = 238

=> x = -17

c) \(\frac{-8}{15}< \frac{x}{40}< \frac{-7}{15}\)

Ta có BCNN(15,40,15) = 120

=> \(\frac{-64}{120}< \frac{3x}{120}< \frac{-56}{120}\)

=> -64 < 3x < -56

=> x \(\in\){ -19;-20;-21}

Câu d tương tự

d) \(\frac{-1}{2}< \frac{x}{18}< \frac{-1}{3}\)

\(\Leftrightarrow\frac{-9}{18}< \frac{x}{18}< \frac{-6}{18}\)

\(\Leftrightarrow-9< x< -6\)

\(\Rightarrow x\in\left\{-8;-7\right\}\)

Theo đầu bài ta có:
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\left(\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+...+\frac{1}{x\left(x+1\right)}\right)\cdot2=\frac{2}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{18}\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)

6. \(\dfrac{x}{4}=\dfrac{9}{x}\)

=>x²=4.9=36

=>x\(\in\)\(\left\{-6;6\right\}\)

\((\dfrac{2x}{5}+2):\left(-4\right)=-1\dfrac{1}{2}\) 

(\(\dfrac{2x}{5}+2):\left(-4\right)=-\dfrac{3}{2}\) 

\(\dfrac{2x}{5}=-\dfrac{3}{2}.\left(-4\right)\) 

\(\dfrac{2x}{5}=6\) 

\(\dfrac{2x}{5}=\dfrac{30}{5}\) 

2x = 30 

x = 30 : 2 = 15

a) 2 - ( \(5\frac{3}{8}\)x  X  - \(\frac{5}{24}\)) = \(\frac{5}{12}\)

             \(5\frac{3}{8}\)x  X  - \(\frac{5}{24}\)= \(\frac{19}{12}\)

             \(5\frac{3}{8}\)x  X             = \(\frac{43}{24}\)

                            X             = \(\frac{1}{3}\)

b)  \(1\frac{2}{9}\): ( \(3\frac{1}{3}\)x  X + \(\frac{1}{6}\))  = \(\frac{22}{23}\)

                    \(3\frac{1}{3}\)x  X + \(\frac{1}{6}\)   = \(\frac{23}{18}\)

                   \(3\frac{1}{3}\)x  X                = \(\frac{10}{9}\)

                                  X                 =\(\frac{1}{3}\)

C)  \(\frac{4}{5}\)x    X  - \(\frac{1}{2}\)x    X + \(\frac{3}{4}\)x    X = \(\frac{7}{40}\)

  (   \(\frac{4}{5}-\frac{1}{2}+\frac{3}{4}\))  x   X                 = \(\frac{7}{40}\)

               \(\frac{21}{20}\)         x   X                      = \(\frac{7}{40}\)

                                       X                       =\(\frac{1}{6}\)

\(x+1⋮x-1\)

\(x-1+2⋮x-1\)

\(2⋮x-1\)hay \(x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(x-2⋮x+1\)

\(x+1-3⋮x+1\)

\(-3⋮x+1\)thự hiện tương tự nhé !