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

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

 mình cảm ơn ạ

a: \(\frac{23}{45}\cdot\frac{15}{16}\cdot\frac{32}{23}+0,3\cdot x=5\frac35\cdot7\frac{7}{12}\cdot\frac{2}{27}\cdot\frac{5}{13}\)

=>\(\frac{12}{16}\cdot\frac{15}{45}+0,3x=\frac{28}{5}\cdot\frac{91}{12}\cdot\frac{2}{27}\cdot\frac{5}{13}\)

=>\(0,3x+\frac{1}{12}=\frac{91}{13}\cdot\frac{2}{12}\cdot\frac{28}{27}=7\cdot\frac16\cdot\frac{28}{27}=\frac{7\cdot14}{3\cdot27}=\frac{98}{81}\)

=>\(0,3x=\frac{98}{81}-\frac{1}{12}=\frac{365}{324}\)

=>\(x=\frac{365}{324}:0,3=\frac{365}{324\cdot0,3}=\frac{365}{97,2}=\frac{1825}{486}\)

b: \(\left\lbrack\left(35\frac57+2\frac34\right)-5\frac57+\frac14\right\rbrack:11+x=3\)

=>\(\left\lbrack35+\frac57+2+\frac34-5-\frac57+\frac14\right\rbrack:11+x=3\)

=>(32+1):11+x=3

=>33/11+x=3

=>x+3=3

=>x=0

( 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

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

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

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

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

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

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

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

Bài 3:

a:

ĐKXĐ: x>=-5

\(x^2-7x=6\sqrt{x+5}-30\)

=>\(x^2-4x-3x+12=6\sqrt{x+5}-18\)

=>\(\left(x-4\right)\left(x-3\right)=6\left(\sqrt{x+5}-3\right)\)

=>\(\left(x-4\right)\left(x-3\right)=6\cdot\frac{x+5-9}{\sqrt{x+5}+3}\)

=>\(\left(x-4\right)\left(x-3-\frac{6}{\sqrt{x+5}+3}\right)=0\)

=>x-4=0

=>x=4(nhận)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{x+4\sqrt{x}+4}=5x+2\)

=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

=>\(5x+2=\sqrt{x}+2\)

=>\(5x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)

b: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)

=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\) =4

=>|x-1|+|x+2|=4(1)

TH1: x<-2

(1) sẽ trở thành: -x-2+1-x=4

=>-2x-1=4

=>-2x=5

=>x=-5/2(nhận)

TH2: -2<=x<1

(1) sẽ trở thành: x+2+1-x=4

=>3=4(vô lý)

TH3: x>=1

(1) sẽ trở thành: x+2+x-1=4

=>2x+1=4

=>2x=3

=>x=3/2(nhận)

c: ĐKXĐ: x>=1

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)

=>\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)

=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\)

=>\(\left|\sqrt{x-1}-1\right|=2-\sqrt{x-1}-1=1-\sqrt{x-1}\)

=>\(\sqrt{x-1}-1\le0\)

=>\(\sqrt{x-1}\le1\)

=>0<=x-1<=1

=>1<=x<=2