\(3^X×3^2=729\:\:\:\:\:\:\\ 5^X×625=3125\\ \left(2X+1\right)^3=27\\ ^{ }\)
Tìm x
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\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)
=>\(\left(\frac{3}{5}\right)^x.\left(\frac{5}{3}\right)^{12}=\left(\frac{3}{5}\right)^5\)
=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{5}{3}\right)^{12}\)
=>\(\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^{17}\)
=>x=17
\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)
\(\Rightarrow\left(\frac{3}{5}\right)^x.\left(\frac{3}{5}\right)^{12}=\left(\frac{3}{5}\right)^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{3}{5}\right)^{12}\)
\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{5}{3}\right)^7\)
\(\Rightarrow x=-7\)
\(4^{x+2}=16=4^2\)
=> x+ 2 = 2
X = 0
\(5^{2x+3}=3125=5^5\)
=> 2x+3 = 5
=> x =1
Mấy phần còn alji tương tự
\(4^{x+2}=16=4^2\)
\(=>x+2=2\) \(=>x=0\)
\(5^{2x+3}=3125=5^5\)
\(=>2x+3=5\) \(=>x=1\)
\(6^{x+2}=216=6^3\)
\(=>x+2=3\) \(=>x=1\)
\(5^{2\left(x+1\right)}=625=5^4\)
\(=>2\left(x+1\right)=4\) \(=>x=1\)
\(3^{4-2x}=3^0\)
\(=>4-2x=0\) \(=>x=2\)
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
Bài 1:
\((1-2x)^2=9=3^2=(-3)^2\)
\(\Rightarrow \left[\begin{matrix} 1-2x=3\\ 1-2x=-3\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-1\\ x=2\end{matrix}\right.\)
Bài 2:
\((x+5)^3=-64=(-4)^3\)
\(\Rightarrow x+5=-4\Rightarrow x=-9\)
Bài 3:
\((3x-5)^2=16=4^2=(-4)^2\)
\(\Rightarrow \left[\begin{matrix} 3x-5=4\\ 3x-5=-4\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=\frac{1}{3}\end{matrix}\right.\)
Bài 4:
\((x-1)^3=27=3^3\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
Bài 5:
\(x^2+x=0\Leftrightarrow x(x+1)=0\)
\(\Rightarrow \left[\begin{matrix} x=0\\ x+1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=0\\ x=-1\end{matrix}\right.\)
Bài 6:
\(5^{x+2}=625=5^4\)
\(\Rightarrow x+2=4\Rightarrow x=2\)
a: \(\Leftrightarrow2x-3=x\)
=>x=3
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+\dfrac{5}{4}\cdot2^x=\dfrac{7}{32}\)
=>2^x=1/8
=>x=-3
c: =>2x+7=-4
=>2x=-11
=>x=-11/2
d: =>(4x-3)^2*(4x-4)(4x-2)=0
hay \(x\in\left\{\dfrac{3}{4};1;\dfrac{1}{2}\right\}\)
Bài làm:
a) Ta có: \(5^{x+2}=625\)
\(\Leftrightarrow5^{x+2}=5^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=2\)
b) \(\left(x-1\right)^{x+2}=\left(-1\right)^{x+4}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}=\left(-1\right)^{x+2}.\left(-1\right)^2\)
\(\Leftrightarrow\left(x-1\right)^{x+2}=\left(-1\right)^{x+2}\)
\(\Rightarrow x-1=-1\)
\(\Rightarrow x=0\)
c) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
5^x+2=625
5^x+2=5^4
x+2=4
x=4-2
x=2
(x-1)^x+2=[(-1)^2]^x+2
(x-1)=(-1)^2
(x-1)=1
x=1+1
x=2
vậy x=2
(2x-1)^3=-8
(2x-1)^3=(-2)^3
2x-1=-2
2x=-2+1
2x=-1
x=-1:2
x=-0,5
vậy x=-0,5
vậy x=2
3^ x . 3^ 2= 729
3^ x . 9 = 729
3^ x = 729: 9
3^x =81
vậy x = 3^3
5^ x . 625 = 3125
5^x = 3125:625
5^x = 5
vậy x = 1
( 2x+1 )^ 3 = 27
( 2x+1) ^ 3 = 3^3
vậy 2x+1 = 3
vậy x = 0