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a, 5^x . (5^3)^2=625
5^x . 5^6=5^4
5^x=5^4:5^6
5^x=5^-2
=> x=-2
a) \(5^x\cdot5^6=5^4\)
\(5^x=5^{4-6}\)
\(5^x=5^{-2}\)
=> x = -2
b) \(\left(5x+1\right)^2=\left(\pm\frac{6}{7}\right)^2\)
+) 5x + 1 = 6/7
5x = -1/7
x = -1/35
+) 5x + 1 = -6/7
5x = -13/7
x = -13/35
Vậy,.........
1: Tìm x
a) Ta có: \(\left(2x-1\right)^3=-27\)
\(\Leftrightarrow2x-1=-3\)
\(\Leftrightarrow2x=-3+1=-2\)
hay x=-1
Vậy: x=-1
b) Ta có: \(\left(2x-3\right)^4=625\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=-5\\2x-3=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-5+3=-2\\2x=5+3=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;4\right\}\)
c) Ta có: \(\left(x-2\right)^5=\left(x-2\right)^7\)
\(\Leftrightarrow\left(x-2\right)^5-\left(x-2\right)^7=0\)
\(\Leftrightarrow\left(x-2\right)^5\left[1-\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)^5\cdot\left[1-\left(x-2\right)\right]\cdot\left[1+\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)^5\cdot\left(1-x+2\right)\cdot\left(1+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)^5\cdot\left(-x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^5=0\\-x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-x=-3\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2;3\right\}\)
d) Ta có: \(5^{x+2}+5^{x+3}=750\)
\(\Leftrightarrow5^{x+2}\cdot1+5^{x+2}\cdot5=750\)
\(\Leftrightarrow5^{x+2}\left(1+5\right)=750\)
\(\Leftrightarrow5^{x+2}\cdot6=750\)
\(\Leftrightarrow5^{x+2}=125\)
\(\Leftrightarrow x+2=3\)
hay x=1
Vậy: x=1
a.(2x +1). (2x+1)=1
Mà chỉ có 1.1=1
Vậy 2x + 1=1
2x=1-1
2x=0
Suy ra: x= 0
Hoàng Khánh Thi thiếu nha.
a) (2x+1)2 = \(\left(\pm1\right)^2\)
=> 2x + 1 = 1 hoặc 2x + 1 = -1
=> 2x = 0 hoặc 2x = -2
=> x = 0 hoặc x = -1.
1. (x2-1/9)+(y-2)2=0
\(\Rightarrow\hept{\begin{cases}x^2-\frac{1}{9}=0\\\left(y-2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2=\frac{1}{9}\\y-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=2\end{cases}}\)
Vậy x = 1/3 và y=2
2. 5x.(53)2=625
=> 5x.56=54
Từ đây ta có thể làm theo 2 cách
C1 : (5x.56=54) = (5x+6 = 54)
<=> x+6 = 4
x = 4-6
x = -2
C2 : 5x.56=54
5x = 54 : 56 (54-6 )
5x = 5-2
<=> x= -2
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{cases}\Rightarrow\orbr{\begin{cases}x=7\\x-7=\pm1\end{cases}}}\)
vậy x=7, x=8 hay x=6
1.
a) 2n : 4 = 16
2n = 16.4
2n = 64
2n = 26
=> n = 6
b) 5n = 625
5n = 54
=> n = 4
c) 2n . 16 = 128
2n = 128 : 16
2n = 8
2n = 23
=> n = 3
d) 3n+3 = 81
3n+3 = 34
=> n + 3 = 4
=> n = 4 - 3
=> n = 1
1.
a) 2n : 4 = 16
2n = 16.4
2n = 64
2n = 26
=> n = 6
b) 5n = 625
5n = 54
=> n = 4
c) 2n . 16 = 128
2n = 128 : 16
2n = 8
2n = 23
=> n = 3
d) 3n+3 = 81
3n+3 = 34
=> n + 3 = 4
=> n = 4 - 3
=> n = 1
2.
a) x3 = 64
x3 = 43
=> x = 4
b) (2x + 1)3 = 27
(2x + 1)3 = 33
2x + 1 = 3
2x = 3 - 1
2x = 2
=> x = 1
c) x3 = x
=> x = {0;1}
d) 25 + 5x = 75.73
25 + 5x = 5764801
5x = 5764801 - 25
5x = 5764776
=> không có số tự nhiên x thoản mãn
Mình k chép lại đề bài nữa nha!
5x . 56 =625
5x+6= 54
=>x+6=4
=> x=-2
vậy x=-2
\(5^x\cdot\left(5^3\right)^2=625\\ \Leftrightarrow5^x\cdot5^6=5^4\\ \Leftrightarrow5^x=\dfrac{5^4}{5^6}\\ \Leftrightarrow5^x=\dfrac{1}{5^2}\\ \Leftrightarrow5^x=5^{-2}\\ \Leftrightarrow x=-2\)
Vậy \(x=-2\)
5x . ( 53 )2 = 625
<=> 5x . 56 = 625
<=> 5x+6 = 625
<=> 5x+6 = 54
<=> x + 6 = 4
<=> x = -2
\(5^x.\left(5^3\right)^2=625\)
\(\Leftrightarrow5^x.5^6=625\)
\(\Leftrightarrow5^x=\frac{1}{5^2}\)
\(\Leftrightarrow x=-2\)