\((85\dfrac{7}{30}-83\dfrac{5}{18}):2\dfrac{2}{3}=0,01x:4\)
Tim x
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b: \(\left[\left(6+\dfrac{3}{5}-3-\dfrac{3}{14}\right)\cdot\dfrac{2}{5}\right]:\left(21-1.25\right)=x:\left(5+\dfrac{5}{6}\right)\)
\(\Leftrightarrow x:\dfrac{35}{6}=\dfrac{237}{175}:\dfrac{79}{4}\)
\(\Leftrightarrow x:\dfrac{35}{6}=\dfrac{12}{175}\)
\(\Leftrightarrow x=\dfrac{2}{5}\)
1: \(\Leftrightarrow\left(\dfrac{x+1}{85}+1\right)+\left(\dfrac{x+3}{83}+1\right)=\left(\dfrac{x+5}{81}+1\right)+\left(\dfrac{x+7}{79}+1\right)\)
=>x+86=0
=>x=-86
2: \(\Leftrightarrow\left(\dfrac{x-1}{2015}+1\right)-\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+7}{2007}+1\right)-\left(\dfrac{x+11}{2003}+1\right)\)
=>x+2014=0
=>x=-2014
3: \(\Leftrightarrow3\left(x+4\right)-2\left(x-3\right)=4x\)
=>4x=3x+12-2x+6
=>4x=x+18
=>3x=18
=>x=6
4: \(\Leftrightarrow15x-5\left(x+1\right)=3\left(2x+1\right)\)
=>15x-5x-5=6x+3
=>10x-5=6x+3
=>4x=8
=>x=2
5: \(\Leftrightarrow2\left(2x-7\right)+5\left(x+11\right)=-40\)
=>4x-14+5x+55=-40
=>9x+41=-40
=>x=-9
Bài 1:
a: \(A=\dfrac{\left(85+\dfrac{7}{30}-83-\dfrac{5}{18}\right):\dfrac{8}{3}}{\dfrac{1}{25}}\)
\(=\left(2+\dfrac{7}{30}-\dfrac{5}{18}\right)\cdot\dfrac{3}{8}\cdot25\)
\(=\dfrac{180+21-25}{90}\cdot\dfrac{75}{8}\)
\(=\dfrac{176}{90}\cdot\dfrac{75}{8}=\dfrac{55}{3}\)
=>12,5% của A là 55/8x1/8=55/64
b: \(B=\dfrac{\left(6+\dfrac{3}{5}-3-\dfrac{3}{14}\right)\cdot\dfrac{36}{5}}{19.75:2.5}\)
\(=\dfrac{\left(3+\dfrac{27}{70}\right)\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{\dfrac{210+27}{70}\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{4266}{175}\cdot\dfrac{10}{79}=\dfrac{108}{35}\)
=>5% là 108/35x1/20=27/175
a: 4/9+3/7=28/63+27/63=55/63
3/4+7/24=18/24+7/24=25/24
1/3+2/9+4/27=9/27+6/27+4/27=19/27
b: 5/6-3/8=20/24-9/24=11/24
7/15-11/30=14/30-11/30=3/30=1/10
2/3+1/6-7/12
=8/12+2/12-7/12
=3/12=1/4
c: 18/25*15/6=15/25*18/6=3*3/5=9/5
30/49:6/7=30/49*7/6=210/294=5/7
1/2*3/4:6/5=3/8*5/6=15/48=5/16
d: 8*3/5:12/5=24/5*5/12=2
4:9/5:10/3=4*5/9*3/10=2/3
\(\dfrac{-2}{45}:\dfrac{8}{3}=0,01x:4\)
\(0,01x:4=\dfrac{-2}{45}\times\dfrac{3}{8}\)
\(0,01x:4=\dfrac{-6}{360}\)
\(0,01x:4=\dfrac{-1}{60}\)
\(0,01x=\dfrac{-1}{60}\times\dfrac{1}{4}\)
\(0,01x=\dfrac{-1}{240}\)
\(x=\dfrac{-1}{240}:0,01\)
\(x=\dfrac{-1}{240}\times100\)
\(x=\dfrac{-100}{240}=-2,4\)
\(\dfrac{-2}{45}:\dfrac{8}{3}=0,01x:4\)
\(\dfrac{-2}{45}\times\dfrac{3}{8}=0,01x:4\)
\(\dfrac{-6}{360}=0,01x:4\)
\(\dfrac{-1}{60}=0,01x:4\)
\(0,01x=\dfrac{-1}{60}\times\dfrac{4}{1}\)
\(0,01x=\dfrac{-4}{60}\)
\(0,01x=\dfrac{-1}{15}\)
\(x=\dfrac{-1}{15}:0,01\)
\(x=\dfrac{-1}{15}\times100\)
\(x=\dfrac{-100}{15}=\dfrac{-20}{3}\)
Vậy \(x=\dfrac{-20}{3}\).
Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)
\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)
\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)
Bài 2:
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)
\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
Vì \(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)
\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow[\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)]:2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow1\dfrac{43}{45}:2\dfrac{2}{3}=0,01x:4\)
\(\Leftrightarrow\dfrac{88}{45}:\dfrac{8}{3}=0,01x:4\)
\(\Leftrightarrow\dfrac{88}{45}.\dfrac{3}{8}=0,01x:4\)
\(\Leftrightarrow\dfrac{11}{15}=0,01x:4\)
\(\Leftrightarrow0,01x:4=\dfrac{11}{15}\)
\(\Leftrightarrow x=\dfrac{11}{15}.4:0,01\)
\(\Leftrightarrow x=\dfrac{880}{3}\)
Vậy x = \(\dfrac{880}{3}\)
Ta có:
\(\left(85\dfrac{7}{30}-83\dfrac{5}{18}\right):2\dfrac{2}{3}=0.01x:4\\ \left(\left(85-83\right)+\left(\dfrac{7}{30}-\dfrac{5}{18}\right)\right):\dfrac{8}{3}=0.01x:4\\ \dfrac{88}{45}:\dfrac{8}{3}=0.01x:4\\ \dfrac{11}{15}=0.01x:4\\ \dfrac{44}{15}=0.01x\\ x=\dfrac{44}{15}:0.01\\ x=\dfrac{880}{3}\)
Vậy \(x=\dfrac{880}{3}\)