Tìm x biết:
(1/12+19/6-30,75).x-8=(3/5+0,415+1/200);0,01
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DP
1
7 tháng 8 2015
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 ^^
TT
3 tháng 5 2017
\(\left(\dfrac{1}{2}+\dfrac{19}{6}-30,75\right).x-8=\left(\dfrac{3}{5}+0,415+\dfrac{1}{200}\right):0.01\)
\(\left(\dfrac{1}{2}+\dfrac{19}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}+\dfrac{1}{200}\right):\dfrac{1}{100}\)
\(\dfrac{-325}{12}.x-8=\dfrac{51}{50}:\dfrac{1}{100}\)
\(\dfrac{-325}{12}.x-8=102\)
\(\dfrac{-325}{12}.x=102+8\)
\(\dfrac{-325}{12}.x=110\)
\(x=110:\dfrac{-325}{12}\)
\(x=\dfrac{-264}{65}\)
2 tháng 7 2023
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
TB
29 tháng 4 2018
\(\left(\dfrac{1}{12}+\dfrac{19}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}+\dfrac{1
}{200}\right)\): \(\dfrac{1}{100}\)
\(\dfrac{-53}{2}\) . 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 = \(\dfrac{-204}{53}\)
30 tháng 5 2017
\((\dfrac{1}{12}+3\dfrac{1}{6}-30,75)x-8=\left(\dfrac{3}{5}+0,415+\dfrac{1}{200}\right):0,01\)
\(\dfrac{-55}{2}.x-8=\dfrac{51}{50}:0,01\)
\(\dfrac{-55}{2}.x-8=102\)
\(\dfrac{-55}{2}.x=110\)
\(x=-4\)
PT
1
CM
20 tháng 2 2018
Ta có:
= (0,6 + 0,415 + 0,005) : 0,01 = 1,02 : 0,01 = 102.
Biểu thức trở thành:
Vậy x = -4.
14 tháng 5 2018
\(M=\left(\frac{3}{5}+0,415\right)+\frac{1}{200}\div0,01\)
\(M=\left(\frac{3}{5}+\frac{83}{200}\right)+\frac{1}{200}\div\frac{1}{100}\)
\(M=\frac{203}{200}+\frac{1}{200}\div\frac{1}{100}\)
\(M=\frac{203}{200}+\frac{1}{2}\)
\(M=\frac{303}{200}\)
\(N=30,75+\frac{1}{12}+3\frac{1}{6}\)
\(N=\frac{123}{4}+\frac{1}{12}+\frac{19}{6}\)
\(N=\frac{185}{6}+\frac{19}{6}\)
\(N=34\)
TT
8 tháng 4 2017
a) \(\left(\dfrac{1}{12}+3\dfrac{1}{6}-30.75\right).x-8=\left(\dfrac{3}{5}+0.415\right)\)
\(=\left(\dfrac{1}{12}+3\dfrac{1}{6}-\dfrac{123}{4}\right).x-8=\left(\dfrac{3}{5}+\dfrac{83}{200}\right)\)
\(=\dfrac{-55}{2}.x-8=\dfrac{203}{200}\)\(=\dfrac{-55}{2}.x=\dfrac{203}{200}+8=\dfrac{1803}{200}\)
\(x=\dfrac{1803}{200}:\dfrac{-55}{2}=\dfrac{-1803}{5500}\)
Q
8 tháng 4 2017
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
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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: \(A\le x\le B\)
\(\dfrac{-13}{9}\le x\le\dfrac{-11}{18}\)
\(\Rightarrow x=-1\)
