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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
\(\left(\frac{3}{5}\right)^x.\left(\frac{625}{81}\right)^3=\frac{243}{3125}\)
\(\Rightarrow\left(\frac{3}{5}\right)^x.\left(\frac{3}{5}\right)^{12}=\left(\frac{3}{5}\right)^5\)
\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^5:\left(\frac{3}{5}\right)^{12}\)
\(\Rightarrow\left(\frac{3}{5}\right)^x=\left(\frac{5}{3}\right)^7\)
\(\Rightarrow x=-7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 25x + 1 . 125x . 625x + 2 = (52)5
=> (52)x + 1 . (53)x . (54)x+ 1 = 510
=> 52x + 2 . 53x . 54x + 8 = 510
=> 2x + 2 + 3x + 4x + 8 = 10
=> 9x + 2 + 8 = 10
=> 9x = 10 - 2 - 8
=> 9x = 0
=> x = 0 : 9
=> x = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
a.\(3x^2-51=-24\)
\(\Rightarrow3x^2=27\)
\(\Rightarrow x^2=9\)
\(\Rightarrow x=3\)
b.\(5x.\left(5^3\right)^2=625\)
\(\Rightarrow5x=5^5:5^{\left(3x2\right)}\)
\(\Rightarrow5x=5^5:5^6\)
\(\Rightarrow5x=5^{-1}=0,2\)
\(\Rightarrow x=0,2:5\)
\(\Rightarrow x=0,04\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
\(5^x.\left(5^3\right)^2=625=5^4\)
\(5^x=5^4:5^6=5^{-2}\)
Vậy x = -2