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\(\left(x-5\right)^2-1=80\)
\(\left(x-5\right)^2=80+1\)
\(\left(x-5\right)^2=81\)
\(\left(x-5\right)^2=9^2\)
\(\left(x-5\right)=9\)
\(x=9+5\)
\(x=14\)
Vậy ...
\(\left(x-5\right)^2-1=80\)
\(\left(x-5\right)^2=80+1\)\(=81\)
\(\left(x-5\right)^2=9^2\)
\(\left(x-5\right)=9\)
\(x=9+5=14\)
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: Ta có: \(20:\left(x+1\right)=\left(5^2+1\right):13\)
\(\Leftrightarrow x+1=10\)
hay x=9
b: Ta có: \(320:\left(x-1\right)=2^2\cdot5^2-20\)
\(\Leftrightarrow x-1=4\)
hay x=5
\(\frac{x}{y}=\frac{-5}{19}+1\)
\(\frac{x}{y}=\frac{14}{19}\)
\(=>x=14;y=19\)
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