Bài 1: 

a) Ta có: \(x\left(x^2-4\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;2;-2\right\}\)

b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)

\(\Leftrightarrow2x-3-3x-x+5=40\)

\(\Leftrightarrow-2x+2=40\)

\(\Leftrightarrow-2x=38\)

hay x=-19

Vậy: x=-19

Bài 2: 

a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)

\(=-45\cdot\left(12+34+54\right)\)

\(=-45\cdot100\)

\(=-4500\)

b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)

\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)

\(=43\cdot57-33\cdot57\)

\(=57\cdot\left(43-33\right)\)

\(=57\cdot10=570\)

|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)

|7 - \(\dfrac{3}{4}x\)|  - \(\dfrac{3}{2}\) = 2

|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)

|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)

\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\) 

\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)

 5  - |\(x-3\)| = 5

       |\(x-3\)| = 5 - 5

      |\(x-3\)| = 0

      \(x-3\) = 0 

      \(x\) = 3

|\(\dfrac{7}{2}\) - 42| + 14 = 14 ( vô lý xem lại đề bài nhé em)

\(\dfrac{2}{3}\) - |\(x+2\)| = 2

     |\(x+2\)| = \(\dfrac{2}{3}\) - 2

    |\(x+2\)| = - \(\dfrac{4}{3}\)  

   |\(x+2\)| ≥ 0 ∀ \(x\)

Vậy \(x\in\) \(\varnothing\)

c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)

\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)

d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)

Tương tự với a ; b 

a: =>1/3*4+1/4*5+...+1/x(x+1)=10/39

=>1/3-1/4+...+1/x-1/x+1=10/39

=>1/3-1/(x+1)=10/39

=>1/(x+1)=13/39-10/39=3/39=1/13

=>x+1=13

=>x=12

b: =>15x=150

=>x=10

c: =>120-5x=45

=>5x=75

=>x=15

( x+ 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 40

x + ( 1 + 2 + 3 + 4 + 5 ) = 40

x + 15 = 40

x = 25

( x+ 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 40

x + ( 1 + 2 + 3 + 4 + 5 ) = 40

x + 15 = 40

x = 25

( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 40

=> x + ( 1 + 2 + 3 + 4 + 5 ) = 40

=>  x + 15 = 40

=> x = 40 - 15 

=> x = 25

Chúc bn học giỏi ^^

\(a,x+\frac{4}{5}-x+4=\frac{x}{3}-x-1\)

\(x+\frac{24}{5}-x=\frac{x}{3}-x-1\)

\(x+\frac{24}{5}-x-\frac{x}{3}+x+1=0\)

\(x+\frac{29}{5}-\frac{x}{3}=0\)

\(x-\frac{1}{3}x=-\frac{29}{5}\)

\(\frac{2}{3}x=-\frac{29}{5}\)

\(x=-\frac{87}{10}\)