x + 20 = 190

=> x= 90 - 20 = 70

x+20=80+110

x+20=190

x=190-20

x=170

x + 20 = 80 + 110

x + 20 = 190

x = 190 - 20

x = 170

\(240-\left(x+5\right)=137\)

\(x+5=240-137\)

\(x+5=103\)

\(x=103-5=98\)

b) \(\left(2x-10\right)-20=110\)

\(2x-10=110+20\)

\(2x-10=130\)

\(2x=130+10\)

\(2x=140\)

\(x=140:2=70\)

\(a,240-\left(x+5\right)=137\\ x+5=103\\ x=98\\ b,\left(2x-10\right)-20=110\\ 2x-10=130\\ 2x=140\\ x=70\)

a, 240-(x+5)=137
            x+5=240-137

            x+5=103

              x=   103-5=98

b,(2x-10)-20=110

     2x-10=110+20

      2x-10=130

      2x=120

         x=120:2

          x=60       

a)X=100

b)X=70

K MKNHA BN

x + 20 - 10 = 110

x + 20 = 110 + 10

x + 20 = 120

x = 120 - 20

x = 100

x + 80 - 50 = 100

x + 80 = 100 + 50

x + 80 = 150

x = 150 - 80

x = 70

x+ 20= 110+10

x+20= 120

      x= 120-20

      x= 100

x+80= 100+50

x+80= 150

      x= 150-80

      x=70 

k nhé

\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...\left|x+\frac{1}{110}\right|=11x\)

\(\Leftrightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+\left|x+\frac{1}{12}\right|+...\left|x+\frac{1}{110}\right|\ge0\)

\(\rightarrow11x\ge0\rightarrow x\ge0\)

\(\text{Ta có:}\)

\(x+\frac{1}{2}+...+x+\frac{1}{110}=11x\)

\(\rightarrow10x+\frac{10}{11}=11x\)

\(\rightarrow x=\frac{10}{11}\)

Sửa đề

\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)

\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)

\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)

\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)

Ko đề cho thêm \(\dfrac{20}{4²}\) mà 

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{11-10}{10.11}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(=1-\frac{1}{11}=\frac{10}{11}\)

Phương trình ban đầu tương đương với: 

\(10x+\frac{10}{11}=11x\)

\(\Leftrightarrow x=\frac{10}{11}\)

\(\Leftrightarrow-2x+\dfrac{3}{20}=1-\dfrac{1}{2}+1-\dfrac{1}{6}+...+1-\dfrac{1}{110}\)

\(\Leftrightarrow-2x+\dfrac{3}{20}=10-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)

\(\Leftrightarrow-2x+\dfrac{3}{20}=10-\dfrac{10}{11}=\dfrac{100}{11}\)

=>-2x=1967/220

hay x=-1967/440

a) 120 - 20 : x = 110

20:x= 120 - 110

20:x= 10

x= 20:10=2

b) 213 - ( x - 13 ) = 1442 : 14

213 - (x-13)=103

x-13= 213 - 103

x-13=110

x=110+13

x=123

c) 2550 : [ 120 - ( x - 5 ) ] = 25

120 - (x-5) = 2550:25

120 - (x-5)=102

x-5 = 120 - 102

x-5= 18

x=18+5

x=23

a,120-20:X=110

20:X=120-110

20:X=10

     X=20:10

     X=2

 b, 213-[X-13]=1442:14

213-[x-13]=103

        x-13=213-103

         x-13=110

          x=110+13

           x=123

c,2550:[120-{x-5}]=25

120-{x-5}=2550:25

120-x-5=102

       x-5=120-102

       x-5=18

        x=18+5

         x=23