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a) 2x = 16 b) 3x + 1 = 9x
2x = 24 3x + 1 = 32x
x = 4 x + 1 = 2x
x = 1
c) 23x + 2 = 4x + 2
23x + 2 = 22(x + 2)
3x + 2 = 2(x + 2)
3x + 2 = 2x + 4
x = 2
d) 32x - 1 = 243
32x - 1 = 35
2x - 1 = 5
2x = 6
x = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a)(2x-3)2=16
=>2x-3=4 hoặc 2x-3=-4
<=>2x=7 hoặc 2x=-1
<=>x=7/2 hoặc x=-1/2
b)(3x-2)5=243=35
=>3x-2=3
=>3x=5
=>x=5/3
c)(7x+2)-1=52
<=>\(\frac{1}{7x+2}=25\)
<=>25(7x+2)=1
<=>175x+50=1
<=>175x=-49
<=>x=-49:175
<=>x=-7/25
d)(x-3/4)4=81=34=(-3)4
=>x-3/4=3 hoặc x-3/4=-3
<=>x=3+3/4 hoặc x=-3+3/4
<=>x=15/4 hoặc x=-9/4
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(3^{x+1}=9^x\)
\(3^{x+1}=\left(3^2\right)^x\) c)
\(3^{x+1}=3^{2x}\)
\(\Rightarrow x+1=2x\)
\(1=2x-x\)
\(1=x\)
Vậy x=1
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(2x - 3)2=16
(2x-3)2=42
=>2x-3=4 hoặc 2x-3=-4
=>2x=7 hoặc 2x=-1
=>x=7/2 hoặc x=-1/2
( 3x - 2 )5 = -243 đề đúng phải là ( 3x - 2 )5=-243
![](https://rs.olm.vn/images/avt/0.png?1311)
a. (x - 2)2 = 1
<=> (x - 2)2 = 12 = (-1)2
<=> \(\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\begin{cases}x=3\\x=1\end{cases}\)
Vậy x \(\in\){1; 3}.
b. (2x - 1)3 = -8
<=> (2x - 1)3 = (-2)3
<=> 2x - 1 = -2
<=> 2x = -2 + 1
<=> 2x = -1
<=> x = -1/2
Vậy x = -1/2.
c. (x + 1/2)2 = 1/16
<=> (x + 1/2)2 = (1/4)2 = (-1/4)2
<=> \(\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Leftrightarrow\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}\)
Vậy x \(\in\){-1/4; -3/4}.
d. (x - 2)3 = -27
<=> (x - 2)3 = (-3)3
<=> x - 2 = -3
<=> x = -3 + 2
<=> x = -1
Vậy x = -1.
a.\(\left(x-2\right)^2\)=1
<=> x-2=1 hoặc x-2=-1
<=> x= 3 hoặc x=1
b.\(\left(2x-1\right)^3\)=-8
\(\left(2x-1\right)^3\)=\(\left(-2\right)^3\)
2x-1=-2
2x=-1
x=-1/2
c.\(\left(x+\frac{1}{2}\right)^2\)=\(\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2\)=\(\left(\frac{1}{4}\right)^2\)hoặc \(\left(x+\frac{1}{2}\right)^2\)=\(\left(-\frac{1}{4}\right)^2\)
x+\(\frac{1}{2}\)=\(\frac{1}{4}\) hoặc x+\(\frac{1}{2}\)=-\(\frac{1}{4}\)
x=-\(\frac{1}{4}\)hoặc x=-\(\frac{3}{4}\)
d.\(\left(x-2\right)^3\)=-27
\(\left(x-2\right)^3\)=\(\left(-3\right)^3\)
x-2=-3
x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(3^{x+1}=9^x=3^{2x}\)
\(\Rightarrow x+1=2x\Leftrightarrow x=1\)
c) \(2^{3x+2}=4^x+5\Leftrightarrow4^{2x+1}=4^{x+5}\)
\(\Rightarrow2x+1=x+5\)\(\Rightarrow x=4\)
d) \(3^{2x-1}=243=3^5\)
\(\Rightarrow2x-1=5\Rightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)3^x+1=9^x
3^x+1=3.3^x
3^x+1=3^x+1
=>x thuộc TH Z
b)2^3.x+2=4^x+5
2^3x+2=2^2.(x+5)
2^3x+2=2^2x+10
2^3x=2^2x+8
3x-2x=8
=>x=8
c)3^2x-1=243
3^2x=243.3
3^2x=729
3^2x=3^6
=>2x=6
x=6:2=3
chúc bạn học tốt nha
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\left(3x-2\right)^5=-243\)
\(\Rightarrow\left(3x-2\right)^5=\left(-3\right)^5\)
\(\Rightarrow3x-2=-3\Rightarrow x=\dfrac{-1}{3}\)
c) Vì \(\left(2x-5\right)^{2000}\ge0\forall x;\left(3y+4\right)^{2002}\ge0\forall y\)
\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\forall x,y\)
Mà theo bài ra \(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)
\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right........\)
(2x - 3)2 = 16
=> (2x - 3)2 = 42
=> \(\orbr{\begin{cases}2x-3=4\\2x-3=-4\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)
(3x - 2)5 = -243
=> (3x - 2)5 = -35
=> 3x - 2 = -3
=> 3x = - 3 + 2
=> 3x = -1
=> x = -1/3
a/ \(\left(2x-3\right)^2=4^2\Rightarrow2x-3=4\Rightarrow x=\frac{7}{2}.\)
b/ \(\left(3x-2\right)^5=\sqrt[5]{-243}^5\Rightarrow3x-2=\sqrt[5]{-243}\)
\(\Rightarrow x=\frac{\sqrt[5]{-243}+2}{3}\)