a) 5.5x = 625
=> 5x+1 = 54
=> x + 1 = 4
=> x = 4 - 1
=> x = 3
Vậy x = 3
b) 8(x + 25) - 155 = 181
=> 8(x + 25) = 181 + 155
=> 8(x + 25) = 336
=> x + 25 = 336 : 8
=> x + 25 = 42
=> x = 42 - 25
=> x = 17
Vậy x = 17
c) (x - 9)4 = (x - 9)2
=> (x - 9)4 - (x - 9)2 = 0
=> (x - 9)2.[(x - 9)2 - 1] = 0
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-9\right)^2=0\\\left(x-9\right)^2-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\\left(x-9\right)^2=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\x-9=1\\x-9=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)
a)5*5x=625
\(\Rightarrow5^{x+1}=625\)
\(\Rightarrow5^{x+1}=5^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
b)8(x+25)-155 =181
\(\Rightarrow8\left(x+25\right)=336\)
\(\Rightarrow x+25=42\)
\(\Rightarrow x=17\)
c) ( x-9)4 = ( x-9)2
\(\Rightarrow\left(x-9\right)^4-\left(x-9\right)^2=0\)
\(\Rightarrow\left(x-9\right)^2\left[\left(x-9\right)^2-1\right]=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-9\right)^2=0\\\left(x-9\right)^2-1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\x-9=\pm1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)
\(a,5.5^x=625\Rightarrow5^x=125\Rightarrow5^x=5^3\Rightarrow x=3\)
\(\text{b) 8(x+25)-155 =181}\Rightarrow8\left(x+25\right)=336\Rightarrow x+25=42\Rightarrow x=17\)
cái còn lại tương tự thôi bạn
ê mấy bn giải các bài kia hộ mk đi
a)5.5^x=625
5^x= 625 : 5
5^x=125
5^x=5^3
b)8(x+25)-155 = 181
8(x+25) = 181+155
8(x+25)=336.
x+25 = 336 : 8
x+25=42
x=42-25
x=17.
a) \(5.5^x=625\)
\(\Rightarrow5^x=625:5\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
b) \(8.\left(x+25\right)-155=181\)
\(\Rightarrow8.\left(x+25\right)=336\)
\(\Rightarrow x+25=42\)
\(\Rightarrow x=17\)
Vậy x = 17
c) \(\left(x-9\right)^4=\left(x-9\right)^2\)
\(\Rightarrow\left(x-9\right)^4-\left(x-9\right)^2=0\)
\(\Rightarrow\left(x-9\right)^2.\left[\left(x-9\right)^2-1\right]=0\)
\(\Rightarrow\left(x-9\right)^2=0\) hoặc \(\left(x-9\right)^2-1=0\)
+) \(\left(x-9\right)^2=0\)
\(\Rightarrow x-9=0\)
\(\Rightarrow x=9\)
+) \(\left(x-9\right)^2-1=0\)
\(\Rightarrow\left(x-9\right)^2=1\)
\(\Rightarrow x-9=\pm1\)
+ \(x-9=1\Rightarrow x=10\)
+ \(x-9=-1\Rightarrow x=8\)
Vậy \(x\in\left\{9;10;8\right\}\)