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\(a,50\%x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)
\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{3}{2}x=1\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)
\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)
\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{25}{4}\)
\(\dfrac{2x}{14}-\dfrac{x}{14}=-\dfrac{21}{14}\)
\(2x-x=-21\)
\(x=-21\)
a) Ta có: 2x+33=-11
nên 2x=-44
hay x=-22
b) Ta có: \(\dfrac{x}{2}=\dfrac{-49}{14}\)
nên x=-7
c) Ta có: \(\dfrac{5}{6}x+\dfrac{10}{3}=\dfrac{7}{2}\)
nên \(\dfrac{5}{6}x=\dfrac{7}{2}-\dfrac{10}{3}=\dfrac{1}{6}\)
hay \(x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\)
\(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\left(x\ne5;-2\right).\\ \Leftrightarrow\dfrac{3}{x-5}+\dfrac{4}{x+2}=0.\\ \Leftrightarrow\dfrac{3x+6+4x-20}{\left(x-5\right)\left(x+2\right)}=0.\\ \Rightarrow7x=14.\\ \Leftrightarrow x=2\left(TM\right).\)
a: \(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{3;1;7;-3\right\}\)
\(\dfrac{6^x}{3^3\cdot4}=2\)
=>6^x=2*4*3^3=8*3^3=6^3
=>x=3
(1 + 2 + 3 + ... + 31 + 32) : 11 + 2
= [(32 + 1).(32 - 1 + 1) : 2] : 11 + 2
= (33 . 32 : 2) : 11 + 2
= 33 . 16 : 11 + 2
= 3 . 16 + 2
= 48 + 2
= 50
\(=3\cdot25-27:9+25\cdot4-18:9=75-3+100-2=72+100-2=170\)
\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{1}{2}\)
\(\dfrac{1}{3}:x=\dfrac{1}{2}-\dfrac{2}{3}\)
\(\dfrac{1}{3}:x=-\dfrac{1}{6}\)
\(x=\dfrac{1}{3}:\left(-\dfrac{1}{6}\right)\)
\(x=-2\)
Vậy ...
#AvoidMe
`2/3 +1/3 : x=1/2`
`=> 1/3 : x= 1/2 -2/3`
`=> 1/3 : x= 3/6 -4/6`
`=> 1/3 : x=-1/6`
`=> x=1/3 :(-1/6)`
`=> x=1/3 xx (-6)`
`=> x= -2`
Vậy `x=-2`