Tìm X Biêt
A, (x-5)^2-1=80
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x/5 + x2 + x/5:1/2=13
x/5 + x.2 + x/5 x 2 = 13
x/5.(1 + 2 ) + 2x = 13
x/5 .3 +2x = 13
3x/5 + 2x = 13
3x + 10x = 65
13x = 65
x = 65 : 13
x = 5
a) \(\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
\(x=\frac{\left(\frac{4}{5}-\frac{1}{2}\right)}{\frac{2}{3}}\)
\(x=\frac{9}{20}\)
b) \(\left|x+\frac{3}{4}\right|-\frac{1}{2}=0\)
\(\left|x+\frac{3}{4}\right|=0+\frac{1}{2}\)
\(\left|x+\frac{3}{4}\right|=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{-1}{4}\\x=\frac{-5}{4}\end{cases}}}\)
Vậy x=-1/4 hoặc x=-5/4
c) \(\left(x+\frac{1}{3}\right)^3=\frac{-1}{8}\)
\(\Leftrightarrow x+\frac{1}{3}=\frac{-1}{8}=\frac{\left(-1\right)^3}{2^3}=\frac{-1}{2}\)
\(x=\frac{-1}{2}-\frac{1}{3}\)
\(x=\frac{-5}{6}\)
\(\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
\(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}\)
\(\frac{2}{3}x=\frac{3}{10}\)
\(x=\frac{3}{10}:\frac{2}{3}\)
\(x=\frac{9}{20}\)
b) l x + 3/4 l - 1/2 = 0
l x + 3/4 l = 1/2
TH1 : \(x+\frac{3}{4}\le0\) TH2: \(x+\frac{3}{4}\ge0\)
=> \(x+\frac{3}{4}=-\frac{1}{2}\) => \(x+\frac{3}{4}=\frac{1}{2}\)
\(x=-\frac{1}{2}-\frac{3}{4}\) \(x=\frac{1}{2}-\frac{3}{4}\)
\(x=-\frac{5}{4}\) \(x=-\frac{1}{4}\)
c) ( x + 1/3 )3 = ( -1/8 )
( x + 1/3 ) 3 = ( -1/3 )3
=> x + 1/3 = -1/3
x = -1/3 - 1/3
x = -2/3
a) [ 2x ] = -1
=> 2x = 1
x = 1 : 2
x = 0,5
b ) [ x + 0.4 ] = 3
=> [ x + 0 ] = 3
=> x + 0 = 3 hoặc x + 0 = -3
+) x + 0 = 3
x = 3 - 0
x = 3
+) x + 0 = -3
x = ( -3 ) -0
x = -3
=> x thuộc { -3 ; 3 }
a) x = (-1) : 2 = -0,5
b) x = 3 - 0,4 = 2,6
c) x = (3 + 5) : 2/3 = 12
a) \(\left(x+5\right)^2=100\Leftrightarrow\orbr{\begin{cases}\left(x+5\right)^2=10^2\\\left(x+5\right)^2=\left(-10\right)^2\end{cases}\Leftrightarrow\orbr{\begin{cases}x+5=10\\x+5=-10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=-15\end{cases}}}\)
b) \(\left(2x-4\right)^2=0\Leftrightarrow2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
c) \(\left(x-1\right)^3=27\Leftrightarrow\left(x-1\right)^3=3^3\Leftrightarrow x-1=3\Leftrightarrow x=4\)
a) \(\left(x+5\right)^2=100\)
=> \(\orbr{\begin{cases}\left(x+5\right)^2=10^2\\\left(x+5\right)^2=\left(-10\right)^2\end{cases}}\)
=> \(\orbr{\begin{cases}x+5=10\\x+5=-10\end{cases}}\)
=> \(\orbr{\begin{cases}x=5\\x=-15\end{cases}}\)
b) \(\left(2x-4\right)^2=0\)
=> \(2x-4=0\)
=> \(2x=4\)
=> \(x=2\)
c) \(\left(x-1\right)^3=27\)
=> \(\left(x-1\right)^3=3^3\)
=> \(x-1=3\)
=> \(x=4\)
\(a\times5:6=25\)
\(a\times5=25\times6\)
\(a\times5=150\)
\(a=150:5\)
\(a=30\)
\(a\times5:6=25\\ a\times\dfrac{5}{6}=25\\ a=25:\dfrac{5}{6}\\ a=25\times\dfrac{6}{5}\\ a=30\)
\(\left(x-5\right)^2-1=80\)
\(\left(x-5\right)^2=80+1\)
\(\left(x-5\right)^2=81\)
\(\left(x-5\right)^2=9^2\)
\(\left(x-5\right)=9\)
\(x=9+5\)
\(x=14\)
Vậy ...
\(\left(x-5\right)^2-1=80\)
\(\left(x-5\right)^2=80+1\)\(=81\)
\(\left(x-5\right)^2=9^2\)
\(\left(x-5\right)=9\)
\(x=9+5=14\)