Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a.\(3^{x-1}=243\)
\(3^x:3^1=243\)
\(3^x=729\)
\(\Leftrightarrow3^6=729\)
\(\Leftrightarrow x=6\)
b.\(\left(\dfrac{2}{3}\right)^{x+1}=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x.\left(\dfrac{2}{3}\right)=\dfrac{8}{4}\)
\(\left(\dfrac{2}{3}\right)^x=3\)
Câu b tính đến đây rồi không mò đc x nữa.
Câu c:
3\(^{x}\) + 3\(^{x+2}\) = 270
3\(^{x}\).(1 + 3\(^2\)) = 270
3\(^{x}\).(1+ 9) = 270
3\(^{x}\).10 = 270
3\(^{x}\) = 270 : 10
3\(^{x}\) = 27
3\(^{x}\) = 3\(^3\)
\(x=3\)
Vậy \(x=3\)
1)(x-3)2=0
=>x-3=0
x=0+3
x=3
2)(2x+1)2=4=22
=>2x+1=2
2x=2-1
2x=1
x=1/2
3)(2x-3)3=8=23
=>2x-3=2
2x=2+3
2x=5
x=5/2
4)(x+1/2)4=1/16=(1/4)4
=>x+1/2=1/4
x=1/4-1/2=1/4-2/4
x=-1/4
5)9,27<=3x<=243
32<=3x<=35 (vì x thuộc Z nên làm tròn số 9,27)
=>x thuộc{3;4}
1) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2=1^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
2) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-2+1\)
\(\Leftrightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
3) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
\(\Leftrightarrow x+2=x+4\)
\(\Leftrightarrow2=4\)
\(\Rightarrow x\in\varnothing\)
4) \(\left(2x-3\right)^5=-243\)
\(\Leftrightarrow\left(2x-3\right)^5=\left(-3\right)^5\)
\(\Leftrightarrow2x-3=-3\)
\(\Leftrightarrow2x=-3+3\)
\(\Leftrightarrow2x=0\)
\(\Rightarrow x=0\)
\(\left(x-2\right)^2=1\)
\(=>\left(x-2\right).\left(x-2\right)=1\)
\(+TH1\)\(\left(x-2\right)=1\)
\(=>1.1=1\left(TM\right)\)
\(+TH2\)\(\left(x-2\right)=-1\)
\(=>\left(-1\right).\left(-1\right)=1\left(TM\right)\)
\(=>x=1;-1\)
bài 1 :
b) (x-1/2 )2 = 0
<=> x - 1/2 = 0
<=> x = 0+ 1/2
<=> x = 1/2
c) ( x - 2 ) 2 = 1
<=> x -2 = 1
<=> x = 1 +2 = 3
d) ( 2x -1 )3 = -8
<=> ( 2x - 1) 3 = ( -2 ) 3
<=> 2x - 1 = -2
<=> 2x = -2+1 = -1
<=> x = -1/2
Bài 2 :
c) 32x-1=243
<=> 32x-1= 35
<=> 2x-1 = 5
<=> 2x = 6
<=> x = 6:2 = 3
Mk chỉ giải đc như vậy thôi
bạn thông cảm nhé !
1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
\(\left(x-2\right)^2=1\)
=> \(\left(x-2\right)^2=1^2\)
=> \(\left(x-2\right)^2=\left(-1\right)^2\)
=> \(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1+2\\x=\left(-1\right)+2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{3;1\right\}.\)
\(\left(2x-1\right)^3=-8\)
=> \(\left(2x-1\right)^3=\left(-2\right)^3\)
=> \(2x-1=-2\)
=> \(2x=\left(-2\right)+1\)
=> \(2x=-1\)
=> \(x=\left(-1\right):2\)
=> \(x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}.\)
\(\left(2x-3\right)^5=-243\)
=> \(\left(2x-3\right)^5=\left(-3\right)^5\)
=> \(2x-3=-3\)
=> \(2x=\left(-3\right)+3\)
=> \(2x=0\)
=> \(x=0:2\)
=> \(x=0\)
Vậy \(x=0.\)
Chúc bạn học tốt!
a) (x - 2)2 = 1
=> x - 2 = 1 hoặc x - 2 = -1
x = 3 ; x = 1
Vậy x = 3; x = 1
b) (2x - 1)3 = -8
=> 2x - 1 = -2
2x = -1
x = \(\frac{-1}{2}\)
Vậy x = \(\frac{-1}{2}\)
c) (2x - 3)5 = -243
=> (2x - 3)5 = (-3)5
=> 2x - 3 = -3
2x = 0
x = 0
Vậy x = 0