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$(2x - \dfrac13)^5 = \dfrac{32}{243}$
$(2x - \dfrac13)^5 = \dfrac{2^5}{3^5}$
$(2x - \dfrac13)^5 = \left(\dfrac23\right)^5$
$2x - \dfrac13 = \dfrac23$
$2x = 1$
$x = \dfrac12$.
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$(\frac{1}{3})^{2x-1}=\frac{1}{243}=(\frac{1}{3})^5$
$\Rightarrow 2x-1=5$
$\Rightarrow 2x=6$
$\Rightarrow x=3$
\(\left(\dfrac{1}{3}\right)^{2x-1}=\dfrac{1}{243}\)
\(\left(\dfrac{1}{3}\right)^{2x-1}=\left(\dfrac{1}{3}\right)^5\)
=> \(2x-1=5\)
\(2x=6\)
\(x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^{2x-1}=243\)
\(\Rightarrow3^{2x-1}=3^5\)
\(\Rightarrow2x-1=5\)
\(\Rightarrow2x=5+1\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=6:2\)
\(\Rightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(3^2x-1=243\)
\(9x-1=243\)
\(9x=244\)
\(x=\frac{244}{9}\)
b,\(3^x+1=3^{2x}\)
\(x+1=2x\)
\(x-2x=-1\)
\(-1x=-1\)
\(x=1\)
Mình chỉ làm được 2 câu thôi chúc cậu học tốt!
a) \(3^{2x-1}=243\)
\(\Rightarrow3^{2x-1}=3^5\)
\(\Rightarrow2x-1=5\)
\(\Rightarrow2x=5+1\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=6:2\)
\(\Rightarrow x=3\)
Vậy \(x=3.\)
b) \(3^{x+1}=9^x\)
Vậy \(x=1.\)
c) \(\left(\frac{4}{7}-\frac{1}{2}x\right)^3=-\frac{8}{34}\)
Câu này hình như đề bài sai, bạn xem lại nhé.
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(\left(2x-3\right)^5=-243\)
\(\Leftrightarrow\left(2x-3\right)^5=\left(-3\right)^5\)
\(\Leftrightarrow2x-3=-3\)
\(\Leftrightarrow2x=-3+3\)
\(\Leftrightarrow2x=0\)
\(\Rightarrow x=0\)
\(\left(x-2\right)^2=1\)
\(=>\left(x-2\right).\left(x-2\right)=1\)
\(+TH1\)\(\left(x-2\right)=1\)
\(=>1.1=1\left(TM\right)\)
\(+TH2\)\(\left(x-2\right)=-1\)
\(=>\left(-1\right).\left(-1\right)=1\left(TM\right)\)
\(=>x=1;-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$(\frac{-1}{3})^3.x=\frac{1}{81}=(\frac{-1}{3})^4$
$\Rightarrow x=(\frac{-1}{3})^4: (\frac{-1}{3})^3=\frac{-1}{3}$
b.
$2^2.16> 2^x> 4^2$
$\Rightarrow 2^2.2^4> 2^x> (2^2)^2$
$\Rightarrow 2^6> 2^x> 2^4$
$\Rightarrow 6> x> 4$
$\Rightarrow x=5$ (với điều kiện $x$ là số tự nhiên nhé)
c.
$9.27< 3^x< 243$
$3.3^3< 3^x< 3^5$
$\Rightarrow 3^4< 3^x< 3^5$
$\Rightarrow 4< x< 5$
Với $x$ là stn thì không có số nào thỏa mãn.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-2\right)^2=1\)
=> \(\left(x-2\right)^2=1^2\)
=> \(\left(x-2\right)^2=\left(-1\right)^2\)
=> \(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=1+2\\x=\left(-1\right)+2\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
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
3^2x-1=243
<=>3^2x-1=3^5
<=>2x-1=5
<=>2x=5+1
<=>2x=6
<=>x=6:2
<=>x=3
Vậy x = 3.
thế thôi!:D
**** cho mk nha!