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\(x-\dfrac{1}{2}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}+\dfrac{1}{2}\)
\(x=\dfrac{5}{4}\)
\(x+\dfrac{7}{8}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{7}{8}\)
\(x=\dfrac{-1}{8}\)
\(\dfrac{1}{2}\cdot x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{2}+\dfrac{1}{4}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{4}\)
\(x=\dfrac{-1}{4}\div\dfrac{1}{2}\)
\(x=\dfrac{-1}{2}\)
Câu D ko bt
c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)
\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)
d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)
hay \(x=\dfrac{2}{3}\)
a, 2/5 + 3/4 : x = -1/2
3/4 : x = -1/2 - 2/5
3/4 : x = -9/10
x = 3/4 : -9/10
x = -5/6
b, 5/7 - 2/3 . x = 4/5
2/3 . x = 4/5 + 5/7
2/3 . x = 53/35
x = 53/35 : 2/3
x = 159/70
\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)
\(\Leftrightarrow x=-\dfrac{13}{24}\)
\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{6}{5}\)
\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)
\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)
\(=\dfrac{1}{5}\)
a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)
\(x=\dfrac{-3}{8}-\dfrac{1}{6}\)
\(x=\dfrac{-13}{24}\)
vậy x =....
b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)
\(\dfrac{3}{4}-x=\dfrac{17}{12}\)
\(x=\dfrac{3}{4}-\dfrac{17}{12}\)
\(x=\dfrac{-2}{3}\)
vậy x =....
a: =>x-3=9
=>x=12
b: =>10-x=-26
=>x=36
c: =>x:4-1=2
=>x:4=3
=>x=12
d: =>x^2=4
=>x=2 hoặc x=-2
e: =>(x-2)^2=100
=>x-2=10 hoặc x-2=-10
=>x=12 hoặc x=-8
a)4/5+x=2/3
x=2/3-4/5
x=-2/15
b)-5/6-x=2/3
x=-5/6-2/3
x=-3/2
c)1/2x+3/4=-3/10
1/2x=-3/10-3/4
1/2x=-21/20
x=-21/20:1/2
x=-21/10
d)x/3-1/2=1/5
x/3=1/5+1/2
x/3=7/10
10x/30=21/30
10x=21
x=21:10
x=21/10
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
a, đk x khác 0
<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)
b, <=> 36x +252 = -360 <=> x = -17
c. đk x khác -1
<=> (x+1)^2 = 16
TH1 : x + 1 = 4 <=> x = 3 (tm)
TH2 : x + 1 = -4 <=> x = -5 (tm)
d, đk x khác 1/2
<=> (2x-1)^2 = 81
TH1 : 2x - 1 = 9 <=> x = 5 (tm)
TH2 : 2x - 1 = -9 <=> x = -4 (tm)
a: \(\Leftrightarrow x^2=16\)
hay \(x\in\left\{4;-4\right\}\)
b: =>x+7/15=-2/3
=>x+7=-10
hay x=-17
c: \(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)
hay \(x\in\left\{3;-5\right\}\)
c. \(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)-\dfrac{3}{2}=\dfrac{1}{4}\)
\(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{3}{2}=\dfrac{7}{4}\)
\(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{4}:2=\dfrac{7}{8}\)
\(\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}=\dfrac{29}{24}\)
\(x=\dfrac{29}{24}:\dfrac{1}{2}=\dfrac{29}{12}\)
Vậy : ...
d. \(\dfrac{4}{5}-\dfrac{1}{2}x=\dfrac{1}{10}\)
\(-\dfrac{1}{2}x=\dfrac{1}{10}-\dfrac{4}{5}=-\dfrac{7}{10}\)
\(x=-\dfrac{7}{10}:\left(-\dfrac{1}{2}\right)=\dfrac{7}{5}\)
Vậy : ...
c) \(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)-\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{3}{2}\)
\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{6}{4}\)
\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}:2\)
\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}.\dfrac{1}{2}\)
\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{8}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{21}{24}+\dfrac{8}{24}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{29}{24}\)
\(\Rightarrow x=\dfrac{29}{24}:\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{29}{24}.2\)
\(\Rightarrow x=\dfrac{29}{12}\)
d) \(\dfrac{4}{5}-\dfrac{1}{2}x=\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{4}{5}-\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{8}{10}-\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}:\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{7}{10}.2\)
\(\Rightarrow x=\dfrac{7}{5}\)
\(\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\)
TH1: \(x-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{2}\)
\(\Rightarrow x=1\)
TH2: \(x-\dfrac{1}{2}=-\dfrac{1}{2}\)
\(\Rightarrow x=-\dfrac{1}{2}+\dfrac{1}{2}\)
\(\Rightarrow x=0\)