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\(27\cdot5^3\cdot3^3\cdot32^{-1}:125=3^3\cdot5^3\cdot3^3\cdot\dfrac{1}{32}:5^3=\dfrac{3^6}{32}=\dfrac{3^6}{2^5}\)
3^x*5^x-1=224
3^x*5^x/5=224
15^x=224*5
15^x=1120
=>ko tồn tại x thỏa mãn đề bài vị 15^x luôn có tận cùng bằng 5 (x khác 0 ) hoặc 1 ( x=0) ma 1120 co tận cùng bằng 0
b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
\(x^5=32^4\)
\(=>x^5=\left(2^5\right)^4=\left(2^4\right)^5\)
\(=>x=2^4=16\)
h cho mình nha!
a) \(\left(\frac{1}{2}\right)^x=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\)
=> x = 5
b) \(\left(\frac{5}{7}\right)^x=\frac{125}{343}\)
\(\left(\frac{5}{7}\right)^x=\left(\frac{5}{7}\right)^3\)
=> x = 3
\(2^x.2^{x+1}=32\)
\(\Rightarrow2^{2x+1}=2^5\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)\(\Rightarrow x=2\)
Vậy \(x=2\)
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy x = 1/2
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow x-2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
Vậy x = 3 hoặc x = 1
\(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
<=> 2x = -1
<=> x = -0,5
Vậy x = -0,5
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
\(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1+2\\x=-1+2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy\(x\in\left\{3;1\right\}\)
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-1\times2\)
\(x=-2\)
\(x\left(\frac{1}{2}\right)^2=\frac{1}{16}\)
\(x\left(\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x\frac{1}{2}=\frac{1}{4}\\x\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}:\frac{1}{2}\\x=-\frac{1}{4}:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
(x/2-3)5=32
<=>(x/2-3)5=25
<=>x/2-3=2
<=>x=(2+3).2
<=>x=10
!!!!!CHÚC!!MAY!!MẮN!!!!!
\(\left(x-1\right)^5=-32\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)
Suy ra: \(x-1=-2\Leftrightarrow x=-1\).
( x - 1 )5 = ( - 2 )5 => x - 1 = - 2 => x =( - 2 ) + 1 => x = -1