Ta làm vế kia trước:

\(\left(\frac{3}{5}+0,415+\frac{1}{200}\right):0,01=\left(\frac{3}{5}+\frac{83}{200}+\frac{1}{200}\right):0,01\)

\(=\left(\frac{120}{200}+\frac{83}{200}+\frac{1}{200}\right):0,01\)

\(=\frac{204}{200}:\frac{1}{100}=\frac{51}{50}.100=102\)

=>  \(\left(\frac{1}{12}+\frac{19}{6}-30,75\right).x-8=102\)

   \(\left(\frac{1}{12}+\frac{19}{6}-\frac{123}{4}\right).x=102+8=110\)

    \(\left(\frac{1}{12}+\frac{38}{12}-\frac{369}{12}\right).x=110\)

     \(-\frac{330}{12}.x=110\)

     \(-\frac{55}{2}.x=110\)

     \(x=110:-\frac{55}{2}=110.-\frac{2}{55}=-4\)

**** CHO MK NHA , TKS ^^

\(\left(\dfrac{1}{2}+\dfrac{19}{6}-30,75\right).x-8=\left(\dfrac{3}{5}+0,415+\dfrac{1}{200}\right):0.01\)

\(\left(\dfrac{1}{2}+\dfrac{19}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}+\dfrac{1}{200}\right):\dfrac{1}{100}\)

\(\dfrac{-325}{12}.x-8=\dfrac{51}{50}:\dfrac{1}{100}\)

\(\dfrac{-325}{12}.x-8=102\)

\(\dfrac{-325}{12}.x=102+8\)

\(\dfrac{-325}{12}.x=110\)

\(x=110:\dfrac{-325}{12}\)

\(x=\dfrac{-264}{65}\)

a: =>4x-5=0 hoặc 5/4x-2=0

=>x=5/4 hoặc x=2:5/4=2*4/5=8/5

b: =>(1/12+19/6-30,75)*x-8=102

=>-55/2x=110

=>x=-4

\(\left(\dfrac{1}{12}+\dfrac{19}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}+\dfrac{1 }{200}\right)\): \(\dfrac{1}{100}\)

\(\dfrac{-53}{2}\) . x - 8 = \(\dfrac{51}{50}\): \(\dfrac{1}{100}\)

\(\dfrac{-53}{2}.x-8=\)102

\(\dfrac{-53}{2}\).x = 102 + 8

\(\dfrac{-53}{2}\).x = 110

x = 110 : \(\dfrac{-53}{2}\)

x = \(\dfrac{-204}{53}\)

\((\dfrac{1}{12}+3\dfrac{1}{6}-30,75)x-8=\left(\dfrac{3}{5}+0,415+\dfrac{1}{200}\right):0,01\)

\(\dfrac{-55}{2}.x-8=\dfrac{51}{50}:0,01\)

\(\dfrac{-55}{2}.x-8=102\)

\(\dfrac{-55}{2}.x=110\)

\(x=-4\)

co may tinh ma ban

Vậy hộ mink mink k cho tại ngại ấn mà

Nói nhầm

Tại mink ngại ấn nên hộ mink đi

Ta có:

 = (0,6 + 0,415 + 0,005) : 0,01 = 1,02 : 0,01 = 102.

Biểu thức trở thành:

Vậy x = -4.

\(M=\left(\frac{3}{5}+0,415\right)+\frac{1}{200}\div0,01\)

\(M=\left(\frac{3}{5}+\frac{83}{200}\right)+\frac{1}{200}\div\frac{1}{100}\)

\(M=\frac{203}{200}+\frac{1}{200}\div\frac{1}{100}\)

\(M=\frac{203}{200}+\frac{1}{2}\)

\(M=\frac{303}{200}\)

\(N=30,75+\frac{1}{12}+3\frac{1}{6}\)

\(N=\frac{123}{4}+\frac{1}{12}+\frac{19}{6}\)

\(N=\frac{185}{6}+\frac{19}{6}\)

\(N=34\)

a) \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30.75\right).x-8=\left(\dfrac{3}{5}+0.415\right)\)

\(=\left(\dfrac{1}{12}+3\dfrac{1}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}\right)\)

\(=\dfrac{-55}{2}.x-8=\dfrac{203}{200}\)\(=\dfrac{-55}{2}.x=\dfrac{203}{200}+8=\dfrac{1803}{200}\)

\(x=\dfrac{1803}{200}:\dfrac{-55}{2}=\dfrac{-1803}{5500}\)

a, \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{3}{5}+0,415\)

\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x-8=\dfrac{203}{200}\)

\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{203}{200}+8\)

\(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30,75\right).x=\dfrac{1803}{200}\)

\(\left(\dfrac{13}{4}-30,75\right).x=\dfrac{1803}{200}\)

\(\dfrac{-55}{2}.x=\dfrac{1803}{200}\)

\(x=\dfrac{1803}{200}:\dfrac{-55}{2}\)

\(x=\dfrac{-1803}{5500}\)

Nếu là tìm số nguyên thì hình như đề sai rồi bạn
_______________________________________

b, \(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)

Cho \(A=4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\)

\(A=4\dfrac{1}{3}.\dfrac{-1}{3}\)

\(A=\dfrac{13}{3}.\dfrac{-1}{3}\)

\(A=\dfrac{-13}{9}\)

Cho \(B=\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)

\(B=\dfrac{2}{3}.\left(\dfrac{-1}{6}-\dfrac{3}{4}\right)\)

\(B=\dfrac{2}{3}.\dfrac{-11}{12}\)

\(B=\dfrac{-11}{18}\)

Ta có: \(A\le x\le B\)

\(\dfrac{-13}{9}\le x\le\dfrac{-11}{18}\)

\(\Rightarrow x=-1\)