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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
1,-12 + x = 5x - 20
-12 + 20 = 5x - x
8 = 4x
x = 2
2, 7x - 4 = 20 + 3x
7x - 3x = 20 + 4
4x = 24
x = 6
3 , 5x - 7 = -21 - 2x
5x + 2x = -21 + 7
7x = -14
x = -14 : 7 = -2
4 , x + 15 = 20 - 4x
x + 4x = 20 - 15
5x = 5
x = 5 : 5 = 1
5, 17 - x = 7 - 6x
-x + 6x = 7 - 17
5x = -10
x = -10 : 5 = -2
\(\left(x-5\right)^2-1=80\)
\(\left(x-5\right)^2=80+1\)
\(\left(x-5\right)^2=81\)
\(\left(x-5\right)^2=9^2\)
\(\left(x-5\right)=9\)
\(x=9+5\)
\(x=14\)
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) \(\left(x-1\right)^3=27\)
=> \(\left(x-1\right)^3=3^3\)
=> \(x-1=3\)
=> \(x=4\)
\(\frac{x}{y}=\frac{-5}{19}+1\)
\(\frac{x}{y}=\frac{14}{19}\)
\(=>x=14;y=19\)
Bài 3:
a: Ta có: \(3x^2=75\)
\(\Leftrightarrow x^2=25\)
hay \(x\in\left\{5;-5\right\}\)
b: Ta có: \(2x^3=54\)
\(\Leftrightarrow x^3=27\)
hay x=3
Bài 2:
b: Ta có: \(30-3\cdot2^n=24\)
\(\Leftrightarrow3\cdot2^n=6\)
\(\Leftrightarrow2^n=2\)
hay n=1
c: Ta có: \(40-5\cdot2^n=20\)
\(\Leftrightarrow5\cdot2^n=20\)
\(\Leftrightarrow2^n=4\)
hay n=2
d: Ta có: \(3\cdot2^n+2^n=16\)
\(\Leftrightarrow2^n\cdot4=16\)
\(\Leftrightarrow2^n=4\)
hay n=2