(x+3)2-1=0 (x-1)3=343 (x-2)2=4096
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) ( x - 1 )3 = 343
73 = 343
x + 1 = 7
x = 7 - 1
x = 6
b ) ( x - 2 )4 = 4096
84 = 4096
x - 2 = 8
x = 8 + 2
x = 10
\(\left(x-1\right)^3=343=7^3\) => x -1 =7 => x =8
\(\left(x-2\right)^4=4096=8^4=\left(-8\right)^4\)\(\Rightarrow\orbr{\begin{cases}x-2=8\\x-2=-8\end{cases}\Rightarrow\orbr{\begin{cases}x=10\\x=-6\end{cases}}}\)
2.Tìm x, biết
a,(x-1)3=3 (x-1)3=343
=> (x-1)3=73
=> x -1 = 7
=> x = 8
B, (X-2)4=4096
(X-2)4= 84
=> x - 2 = 8
=> x = 10
C,(2x2-13)4=(-5)4
=> 2x2-13 = -5
=> 2x2 = 8
=> x2 = 4
=> x = 2
Study well
a. (x-1)3 = 343
=> (x-1)3 = 73
=> x - 1 = 7
=> x = 7 + 1
=> x = 8
(x-1)3=343 <=> (x-1)3=73 <=> x-1=7 <=> x=8
(x-2)4=4096 <=> (x-2)4=84 <=> x-2=8 <=> x=10
(x-4)2= (x-4)4 <=> x-4=0 <=> x=4
2.b,
\(\left(x-2\right)^4=4096\\ \left(x-2\right)^4=\left(\pm8\right)^4\\ \Rightarrow\left\{{}\begin{matrix}x-2=8\\x-2=-8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\x=-6\end{matrix}\right.\)
Vậy ...
câu c:
\(\left(x-4\right)^2=\left(x-4\right)^4\\ \Rightarrow\left\{{}\begin{matrix}x-4=0\\x-4=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy
a: 3^x=243
=>3^x=3^5
=>x=5
b: 4^x=4096
=>4^x=4^5
=>x=5
c: 5^3-x=25
=>3-x=2
=>x=1
d: =>2x-3=3
=>2x=6
=>x=3
j: =>2^x*8+2^x*2=80
=>2^x=8
=>x=3
a,2^x*16=128
2^x=8
2^x=2^3
suy ra x=3
b,3^x/9=27
3^x/3^2=3^3
3^x=3^5
suy ra x=5
c,(2*x+1)^3=27
(2*x+1)^3=3^3
suy ra 2*x+1=3
suy ra 2x=2
suy ra x=1
d,(x*2)^2=(x*2)^4
suy ra (x*2)^4-(x*2)^2=0
(x*2)^2*((x*2)^2-1)=0
suy ra (x*2)^2=0 hoac (x*2)^2-1=0
+(x*2)^2=0 +(x*2)^2-1=0
suy ra x*2=0 (x*2)^2=1
suy ra x=0 suy ra x*2=1 hoac -1
suy ra x=1/2 hoac -1/2
a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)
b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)
c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)
d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)
f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)
g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)
h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)
i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1=1^{49}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4n = 4096
4n = 212
n = 12
5n = 15625
5n = 56
n = 6
6n+3 = 216
6n+3 = 23.33
6n+3 = 63
n + 3 = 3
\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
\(A=3^x+3^{x+1}+...+3^{x+100}\)
=>\(3A=3^{x+1}+3^{x+2}+...+3^{x+101}\)
=>\(2A=3^{x+101}-3^x\)
=>\(A=\dfrac{3^{x+101}-3^x}{2}\)
=>\(3^{x+101}-3^x=3^{105}-3^4\)
=>x=4
Vì \(0.\left(x-1\right)^3=0\)
Mà \(0.\left(x-1\right)^3=4096\)
=>ko có x nào thỏa mãn