a, 105.(17x-34) = 0

=> 17x-34=0

=> 17x=0+34=34

=>x=34:17=2

=>xϵ{2}

b, 121:11-(4x+5):3=4

=>11-(4x+5)=4.3

=>4x-5=11-12

=>4x-5=-1

=>4x=-1+5

=>4x=4

=>x=4:4=1

=>x∈{1}

a) 105 (17x - 34) =0 b) 121:11-(4x+5):3=4

105(17x - 34) = 1785(x - 2)

1785(x - 2) = 0

1785x = 3570

1785x = 1785 x 2

x= 2

x = 4

a)

\(105\left(17x-34\right)=0\)

\(\Leftrightarrow17x-34=0\)

\(\Leftrightarrow x=\frac{34}{17}=2\)

b)

\(\frac{121}{11}-\frac{\left(4x+5\right)}{3}=4\)

\(\Leftrightarrow\frac{121}{11}-4=\frac{\left(4x+5\right)}{3}\)

\(\Leftrightarrow7=\frac{\left(4x+5\right)}{3}\)

\(\Leftrightarrow21=4x+5\)

\(\Leftrightarrow x=\frac{21-5}{4}=4\)

a: =17-43=-26

b: =-34+34+11-11+105=105

c: \(=64+125\cdot100=12564\)

d: \(=60+\left[7^3-343\right]\cdot2017^{2018}=60\)

\(a,25-(-18)+(-43)\)

=\(25+18+(-43)\)

=\(43+(-43)\)

=\(0\)

\(b,(-34)+11+\left|-34\right|+105\)

=\((-34)+11+34+105\)

=\([(-34)+34]+11+105\)

=\(0+11+105\)

=\(116\)

\(c,64+105+147-47.125\)

=\(64+125+(147-47).125\)

=\(64+125+100.125\)

=\(64+125+12500\)

=\(189+12500\)

=\(12689\)

\(d,360\div12.2+[(11-4)^3-343].2020\)

=\(360\div12.2+(7^3-343).2020\)

=\(360\div12.2+0.2020\)

=\(30.2+0\)

=\(60\)

a)=(-65-35)+(28+172)

=-100+100

=0

b)=34.[-65+(-55)+20)

=34.(-100)

=-3400

c)=168-437-168-105+437

=(168-168)+(-437+437)-105

=0+0-105

-105

a; (-65) + 28 + (-35) + 172 

= -(65 + 35) + (28 + 172)

= - 100 +  200

= 100 

b; 34.(-65) + (-55).34 + 34.20

= 34.(-65 - 55 + 20)

= 34.(-120 + 20)

= 34.(-100)

= -3400

c; 168 - (437 + 168) + (-105 + 437)

= 168 - 437 - 168 - 105 + 437

= (168 - 168) - (437 - 437) - 105

= 0 - 0 - 105

= -105

\(\dfrac{1}{2}-\left(\dfrac{-16}{7}\right)+\dfrac{4}{5}-\left(\dfrac{-4}{9}\right)+\dfrac{11}{45}+\dfrac{19}{34}+\dfrac{19}{105}=\dfrac{3836}{765}\approx5\)

\(=\dfrac{1}{2}+\dfrac{16}{17}+\dfrac{19}{34}+\dfrac{4}{5}+\dfrac{4}{9}-\dfrac{11}{45}+\dfrac{19}{105}\)

\(=\dfrac{17+32+19}{34}+\dfrac{36+20-11}{45}+\dfrac{19}{105}\)

\(=2+1+\dfrac{19}{105}=3+\dfrac{19}{105}=\dfrac{334}{105}\)