d) Ta có: \(-8+\left(-14\right)+34+\left(-12\right)\)

\(=-8-14+34-12\)

\(=-20+20\)

\(=-0\)

e) Ta có: \(\left(-7\right)+\left(-20\right)+57+\left(-30\right)\)

\(=\left(-7+57\right)+\left(-20-30\right)\)

\(=50-50=0\)

f) Ta có: \(300-\left(-180\right)-120-42\)

\(=300+180-120-42\)

\(=258+60=318\)

d, Ta có : \(\left(-8\right)+\left(-14\right)+34+\left(-12\right)\)

\(=-\left(8+12\right)+34-14=-20+20=0\)

e, Ta có : \(\left(-7\right)+\left(-20\right)+57+\left(-30\right)\)

\(=-\left(30+20\right)+57-7=-50+50=0\)

f, Ta có : \(300-\left(-180\right)-120-42=318\)

-45 x 65 - 45 x 30 - 45 x 3

= -45 x (65 + 30 + 3)

= -45 x 98

= -230

a) -5x (-9) = -11\(\Rightarrow-5\chi=\frac{11}{9}\Rightarrow\chi=\frac{11}{9}:\left(-5\right)\Rightarrow\chi=\frac{11}{9}.\frac{-1}{5}\Rightarrow\chi=\frac{-11}{45}\)b) 25 - (x + 5 ) = -415 - ( 15-415)\(\Rightarrow25-\left(\chi+5\right)=-415-15+415\)

\(\Rightarrow25-\left(\chi+5\right)=\left(-415+415\right)-15\)

\(\Rightarrow25-\left(\chi+5\right)=-15\Rightarrow\chi+5=25+15\)

\(\Rightarrow\chi+5=40\Rightarrow\chi=35\)

HTDT

\(a)\)

\(\left(-31\right)+\left(50-19\right)-\left(150-31\right)\)

\(=\left(-31\right)+50-19-150+31\)

\(=\left(-150\right)-19\)

\(=-169\)

\(b)\)

\(25.\left(45-17\right)-45.\left(25-17\right)\)

\(=25.45-25.17-45.25+45.17\)

\(=0\)

\(c)\)

\(\frac{-1}{12}+\frac{4}{3}=\frac{5}{4}\)

\(d)\)

\(3+\frac{-5}{20}+\frac{30}{75}+\frac{-7}{4}\)

\(=\left(\frac{3}{5}+\frac{30}{75}\right)-\left(\frac{5}{20}+\frac{7}{4}\right)\)

\(=1-2\)

\(=-1\)

a) (-31)+(50-19)-(150-31)

= (-31)+50+(-19)-150+(-31)
= (-31)+50-150+(-19)-(-31)
= (-31)+(-100)+12
= -119
b) 25(45-17)-45(25-17)
= 25.45-25.17-45.25-45.17
= 25(45-45)-25(17-17)
= 0
c) -1/12 + 4/3
= -1/12 + 16/12
= 15/12
= 5/4
d) 3/5+(-5)/20+30/75+(-7)/4
= 45/75+30/75+(-5)/20+(-35)/20
= 1+(-2)
= -1

#include <bits/stdc++.h>
using namespace std;
char a;
int main()
{
    cin>>a;
    if (a=='S') cout<<"50";
    else cout<<"600";
    return 0;
}

a) 5 69   :   5 67   –   3 40   :   3 39   =   5 2   –   3 1  = 25 – 3 = 22

b) 155 – [ 2 . ( 30 + 5 – 26 ) . ( 24 : 3 )]

= 155 – [ 2 . 9 . 8 ] = 155 – 144 = 11

c) 37 . 143 + 37 . 57 + 1300

= (37 . 143 + 37.57) + 1300

= 37.(143+ 57) + 1300

= 37. 200 + 1300

= 7400 + 1300

8700.

20 – [30 – (5 – 1)2] = 20 – (30 – 42) = 20 – (30 – 16) = 20 – 14 = 6

\(a.20-\left[30-\left(5-1\right)^2\right]\)                             \(b.75-\left(3.5^2-4.2^3\right)\)

\(=20-\left[30-4^2\right]\)                                       \(=75-\left(75-4.2^3\right)\)

\(=20-14=6\)                                          \(=75-\left(75-32\right)\)

                                                                 \(=75-43=32\)

1. a) 17.85 + 15.17 - 120

= 17.( 85 + 15) - 120

= 17. 100 -120

= 1700 - 120

= 1580

b) 20 - [ 30 - (5 - 1)² ]

= 20 - [ 30 - 4² ]

= 20 - [ 30 - 16 ]

= 20 - 14

= 6

2. a) 70 - 5.(x - 3) = 45

              5.(x - 3) = 70 - 45

              5.(x -3) = 25

                 x - 3 = 25 : 5

                 x - 3 = 5

                 x = 5 + 3

                 x = 8 

b) 10 + 2.x = 4⁵ : 4³

    10 + 2.x = 4²

    10 + 2.x = 16

           2.x = 16 - 10

          2.x = 6

             x = 6 : 2

            x =3

1)a:17.85+15.17-120=17.(85+15)-120=17.100-120=1700-120=1580

b: 20-[30-(5-1)^2 = 20-[30-4^2]=20-(30-16)=20-15=5

2)

a:  70-5(x-3)=45

   5(x-3) = 70-45=25

   x-3 = 25:5=5

   x= 5+3=8

b:  10+2x = 4^5:4^3

   10+2x = 4^2

10+2x = 16

2x = 16-10=6

x = 6:2=3

a: \(61\cdot45+61\cdot23-68\cdot51\)

\(=61\left(45+23\right)-68\cdot51\)

\(=68\cdot61-68\cdot51\)

\(=68\left(61-51\right)=68\cdot10=680\)

b: \(3\cdot5^2-\left(75-4\cdot2^3\right)\)

\(=75-75+4\cdot8\)

\(=4\cdot8=32\)

c: \(36:\left\{2^2\cdot5-\left[30-\left(5-1\right)^2\right]\right\}\)

\(=\dfrac{36}{20-30+4^2}\)

\(=\dfrac{36}{-10+16}=\dfrac{36}{6}=6\)

d: \(\left(12\cdot49-3\cdot2^2\cdot7^2\right):\left(2020\cdot2021\right)\)

\(=\dfrac{\left(12\cdot49-12\cdot49\right)}{2020\cdot2021}=0\)