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\(\frac{\left(-3\right)^x}{81}=-27\)
\(\Leftrightarrow\left(-3\right)^x=\left(-27\right).81\)
\(\Leftrightarrow\left(-3\right)^x=-2187\)
\(\Leftrightarrow\left(-3\right)^x=\left(-3\right)^7\)
\(\Leftrightarrow x=7\)
a) 2(x-1)+3(x-3)=-2 b) x-1/3=x-2/2
2x-2+3x-9=-2 2 (x-1)=3(x-2)
(2x+3x)+(-2-9)=-2 2x-2=3x-6
5x+(-11)=-2 2x-3x=-6+2
5x=-2+11 -1x=-4
5x=9 x=4
x=1,8
Nhớ nha!
a)\(\left(\frac{3}{5}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(x=\left(\frac{3}{7}\right)^7\div\left(\frac{3}{7}\right)^5\)
\(x=\left(\frac{3}{7}\right)^2\)
\(x=\frac{9}{49}\)
Vậy...
b)\(\left(-\frac{1}{3}\right)^3.x=\left(\frac{1}{3}\right)^4\)
\(\left(-\frac{1}{3}\right)^3.x=\left(-\frac{1}{3}\right)^4\)
\(x=\left(-\frac{1}{3}\right)^4\div\left(\frac{-1}{3}\right)^3\)
\(x=-\frac{1}{3}\)
Vậy...
c)\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
=>\(x-\frac{1}{2}=\frac{1}{3}\)
\(x=\frac{1}{3}+\frac{1}{2}\)
\(x=\frac{5}{6}\)
Vậy...
d)\(\left(x+\frac{1}{4}\right)^4=\left(\frac{2}{3}\right)^4\)
=>\(x+\frac{1}{4}=\frac{2}{3}\)
\(x=\frac{2}{3}-\frac{1}{4}\)
\(x=\frac{5}{12}\)
Vậy...
Phù, mãi mới xong, tk cho mk nha bn
a, => |5/3.x| = 1/6
=> 5/3.x = -1/6 hoặc 5/3.x = 1/6
=> x = -1/10 hoặc x = 1/10
Tk mk nha
a) \(\frac{-2^2}{3}.x=\frac{-2^5}{3}\)
\(x=\frac{-2^5}{3}:\frac{-2^2}{3}\)
\(x=\frac{-2^5}{3}.\frac{-3}{2^2}\)
\(x=\frac{2^3}{1}\)
\(x=2^3\)
\(x=8\)
Vậy x = 8
b)\(\frac{-1^3}{3}.x=\frac{1}{81}\)
\(x=\frac{1}{81}:\frac{-1^3}{3}\)
\(x=\frac{1}{81}.\frac{-3}{1^3}\)
\(x=\frac{1}{3^4}.\frac{-3}{1^3}\)
\(x=\frac{1}{3^3}.\frac{-1}{1^2}\)
\(x=\frac{-1}{3^3}\)
\(x=\frac{-1}{27}\)
Vậy \(x=\frac{-1}{27}\)
a) \(\left|x-1,7\right|=2,3\)
\(\Rightarrow\orbr{\begin{cases}x-1,7=2,3\\x-1,7=-2,3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=-0,6\end{cases}}\)
b) \(\left|x+\frac{3}{4}\right|-\frac{1}{3}=0\)
\(\Rightarrow\left|x+\frac{3}{4}\right|=\frac{1}{3}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{3}\\x+\frac{3}{4}=-\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{5}{12}\\x=-\frac{13}{12}\end{cases}}\)
a) \(3^{x+1}=243\)
\(\Leftrightarrow3^{x+1}=3^5\)
\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)
b) \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\Leftrightarrow\left(\frac{1}{2}\right)^{x+1}=\left(\frac{1}{2}\right)^6\)
\(\Leftrightarrow x+1=6\Leftrightarrow x=5\)
c) \(\frac{81}{3x}=9\)
\(\Leftrightarrow3x=9\Leftrightarrow x=3\)
d) \(2^{x+1}+2^{x+2}=192\)
\(\Leftrightarrow2^x.2+2^x.4=192\)
\(\Leftrightarrow2^x.6=192\Leftrightarrow2^x=32\Leftrightarrow x=5\)
e) Ta có : \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}\Rightarrow\left(x-1\right)^{2020}+\left(y+2\right)^{2020}\ge0}\)
Mà \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}}\)
Bài giải
a, \(3^{x+1}=243\)
\(3^{x+1}=3^5\)
\(\Rightarrow\text{ }x+1=5\)
\(\Rightarrow\text{ }x=4\)
b, \(\left(\frac{1}{2}\right)^{x+1}=\frac{1}{64}\)
\(\frac{1}{2^{x+1}}=\frac{1}{2^6}\)
\(2^{x+1}=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
c, \(\frac{81}{3x}=9\)
\(27x=81\)
\(x=3\)
d, \(2^{x+1}+2^{x+2}=192\)
\(2^{x+1}\left(1+2\right)=192\)
\(2^{x+1}\cdot3=192\)
\(2^{x+1}=64=2^6\)
\(\Rightarrow\text{ }x+1=6\)
\(\Rightarrow\text{ }x=5\)
e, \(\left(x-1\right)^{2020}+\left(y+2\right)^{2022}=0\)
Mà \(\hept{\begin{cases}\left(x-1\right)^{2020}\ge0\\\left(y+2\right)^{2022}\ge0\end{cases}}\) với mọi x,y nên \(\hept{\begin{cases}\left(x-1\right)^{2020}=0\\\left(y+2\right)^{2022}=0\end{cases}}\Rightarrow\hept{\begin{cases}x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
\(\Rightarrow\text{ }x=1\text{ ; }y=-2\)
(-3)^x/81=-27
(-3)^x:(-3)^4=(-3)^3
(-3)^x=(-3)^3.(-3)^4
(-3)^x=(-3)^7
suy ra x=7
vậy x=7
\(\frac{\left(-3\right)^x}{81}=-27\)
\(\Rightarrow\left(-3\right)^x=-27.81=-2187\)
\(\Rightarrow\left(-3\right)^x=\left(-3\right)^7\)
\(\Rightarrow x=7\)