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a) x+3/4=7/4
x=7/4-3/4
x=4/4=1
b) 4/5+5/6:x=1/6
5/6:x=1/6-4/5
5/6:x=-19/30
x=-19/30.5/6=-19/25
a) x + 3/4 = 7/4
x = 7/4 - 3/4
x = 4/4 = 1
vậy: x=1
b) 4/5 + 5/6 : x = 1/6
5/6 : x = 1/6 - 4/5
5/6 : x = -19/ 30
x = 5/6 : (-19/30)
x = -25/19
vậy: x= -25/19
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a, \(x\) \(\times\) \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{6}\)
b, \(x\) : \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\) = \(\dfrac{5}{6}\)
\(x\): \(\dfrac{1}{2}\) = \(\dfrac{5}{6}\) + \(\dfrac{3}{4}\)
\(x\) : \(\dfrac{1}{2}\) = \(\dfrac{19}{12}\)
\(x\) = \(\dfrac{19}{12}\) \(\times\) \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{19}{24}\)
c, \(x\) \(\times\) \(\dfrac{3}{4}\) + \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) 1 = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\)
d, \(x\times\) \(\dfrac{3}{4}\) - \(x\) \(\times\) \(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
\(x\) \(\times\) ( \(\dfrac{3}{4}\) - \(\dfrac{1}{4}\)) = \(\dfrac{7}{8}\)
\(x\) \(\times\) \(\dfrac{1}{2}\) = \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{7}{8}\) : \(\dfrac{1}{2}\)
\(x\) = \(\dfrac{7}{4}\)
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( x - 5 )4 = ( x - 5 )6
=> ( x - 5 )6 - ( x - 5 )4 = 0
=> ( x - 5 )4 . [ ( x - 5 )2 - 1 ] = 0
=> ( x - 5 )4 = 0 hoặc ( x - 5 )2 + 1 = 0
TH1 : ( x - 5 )4 = 0 => x - 5 = 0 => x = 5
TH2 : ( x - 5 )2 + 1 = 0
=> ( x - 5 )2 = 1
=> x - 5 = 1 hoặc -1
=> x = 6 hoặc 4
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(x - 5)6 = (x - 5)4
(x - 5)6 - (x - 5)4 = 0
(x - 5)4 . (x - 5)2 - (x - 5)4 . 1 = 0
(x - 5)4 . [(x - 5)2 - 1] = 0
=> (x - 5)4 = 0
=> x - 5 = 0
=> x = 5
Hoặc (x - 5)2 - 1 = 0
=> (x - 5)2 = 1
Vì 2 là số chẵn
=> x - 5 = 1 hoặc x - 5 = -1
=> x = 6 hoặc x = 4
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#)Giải :
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Leftrightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
\(\Leftrightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Leftrightarrow\left(x-5\right)^4.\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\)x=5 hoặc x=6 hoặc x=4