4: =4+5+{12+(-8)-(-8)-(-5)+(-12)+16-5+3}

=9+[12+5-12+16-5+3}

=9+19=28

5: =15-[13+8-10-16]-{5-7+9-3}

=15-(-5)-4

=15+5-4

=20-4

=16

g: =(32-34)+(36-38)+(40-42)

=(-2)+(-2)+(-2)

=-6

l: =-5-{-[7-10]-[5+12]}-[-3+9-4-5]

=-5+(-3)+17-[-3]

=-8+17+3

=20-8=12

\(p,71\cdot70+35\cdot\left(-7\right)-13\cdot35\)

\(=4970+\left(-245\right)-455=4725-455=4270\)

\(q,18\cdot\left(23-17\right)-13\cdot\left(23+18\right)\)

\(=18\cdot6-13\cdot41=108-533=-425\)

m) 63.83-9.7.(-17)

= 5229-63.(-17)

= 5229 - (-1071)

= 5229 + 1071

= 6300

n) 42.3 - 7[(-34)+18]

= 126 - 7. (-16)

= 126 - (-112)

=126+112

= 238

9: =>x-2/3=1/5*-7/15=-7/75

=>x=-7/75+50/75=43/75

11: =>x-4/5=3/10*(-7)/3=-7/10

=>x=-7/10+4/5=8/10-7/10=1/10

12:=>7/5*9+x=4/9

=>x=4/9-63/5=-547/45

10: =>x+5/6=-8/147

=>x=-87/98

\(b,4\dfrac{1}{2}x-\dfrac{1}{2}x=\dfrac{2}{3}:\dfrac{4}{9}\\ \dfrac{9}{2}x-\dfrac{1}{2}x=\dfrac{3}{2}\\ \left(\dfrac{9}{2}-\dfrac{1}{2}\right)x=\dfrac{3}{2}\\ 4x=\dfrac{3}{2}\\ x=\dfrac{3}{2}:4\\ x=\dfrac{3}{8}\)

a) \(-x+80=-220-5x\)

\(\Rightarrow-x+5x=-220-80\)

\(\Rightarrow4x=-300\)

\(\Rightarrow x=-\dfrac{300}{4}\)

\(\Rightarrow x=-75\)

b) \(98+\left(x-12\right)+68=-80-x\)

\(\Rightarrow166+\left(x-12\right)=-80-x\)

\(\Rightarrow166+x-12=-80-x\)

\(\Rightarrow154+x=-80-x\)

\(\Rightarrow154+80=-x-x\)

\(\Rightarrow-2x=234\)

\(\Rightarrow x=-\dfrac{234}{2}\)

\(\Rightarrow x=-117\)

c) \(122+x-78=-55-4x-1\)

\(\Rightarrow44+x=-56-4x\)

\(\Rightarrow x+4x=-56-44\)

\(\Rightarrow5x=-100\)

\(\Rightarrow x=-\dfrac{100}{5}\)

\(\Rightarrow x=-20\)

d) \(663+9x=-x-37\)

\(\Rightarrow9x+x=-37-663\)

\(\Rightarrow10x=-700\)

\(\Rightarrow x=-\dfrac{700}{10}\)

\(\Rightarrow x=-70\)

a) \(\left(2x-4\right)\left(x-2\right)^3=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(x-2\right)^3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=4\\x-2-0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)

\(\Rightarrow x=2\)

b) \(3^{x-1}:81=3^3\)

\(\Rightarrow3^{x-1}:3^4=3^3\)

\(\Rightarrow3^{x-1-4}=3^3\)

\(\Rightarrow3^{x-5}=3^3\)

\(\Rightarrow x-5=3\)

\(\Rightarrow x=8\)

c) \(x^{13}=x\)

\(\Rightarrow x^{13}-x=0\)

\(\Rightarrow x\left(x^{12}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}=1\end{matrix}\right.\)

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

d) \(\left(x-9\right)^4=\left(x-9\right)^2\)

\(\Rightarrow\left(x-9\right)^2=x-9\)

\(\Rightarrow\left(x-9\right)^2-\left(x-9\right)=0\)

\(\Rightarrow\left(x-9\right)\left(x-9-1\right)=0\)

\(\Rightarrow\left(x-9\right)\left(x-10\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x-10=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)

`a,-2022.(x+8)=0`

`=>x+8=0:(-2022)`

`=>x+8=0`

`=>x=0-8`

`=>x=-8`

`b,(7-x)(x+3)=0`

`@ TH1`

`7-x=0`

`=>x=7-0`

`=>x=7`

`@ TH2`

`x+3=0`

`=>x=0-3`

`=>x=-3`

`c,2023x .(14-x)=0`

`@ TH1`

`2023x=0`

`=>x=0:2023`

`=>x=0`

`@ TH2`

`14-x=0`

`=>x=14-0`

`=>x=14`

`d,x^2-x=0`

`=>x(x-1)=0`

`@ TH1`

`x=0`

`@ TH2`

`x-1=0`

`=>x=0+1`

`=>x=1`

\(a,\left(-2022\right).\left(x+8\right)=0\\ \Rightarrow\left(x+8\right)=0:\left(-2022\right)\\ \Rightarrow x+8=0\\ \Rightarrow x=-8\\ b,\left(7-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right. \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\\ c,2023x.\left(14-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2023x=0\\14-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\\ d,x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(\left(-2020\right).\left(x+8\right)=0\)

\(=>x+8=0=>x=-8\)

\(\left(7-x\right).\left(x+3\right)=0\)

\(=>\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

\(2023x.\left(14-x\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\14-x=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\)

\(x^2-x=0\)

\(x\left(x-1\right)=0\)

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