Tìm x biết: 3^2x = 27. 3^x
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\(\dfrac{2x-3}{3}=\dfrac{27}{2x-3}\)
=>(2x-3)2=27.3
2x-3=9 hoặc 2x-3= -9
x=6 hoặc x=-3
\(\dfrac{2x-3}{3}=\dfrac{27}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ \Leftrightarrow\left(2x-3\right)^2=27.3=81\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=9\\2x-3=-9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\left(tmđk\right)\\x=-3\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x\in\left\{6;-3\right\}\)
x^3 = 27
x^3 = 3^3
=> x = 3
(2x-1)^3 = 8
(2x-1)^3 = 2^3
2x-1 = 2
2x = 2+1
2x = 3
x = 3:2
=> ko có x phù hợp.
(x-2)^2 = 16
(x-2)^2 = 4^2
x-2 = 4
x = 4+2
x = 6
Chúc bn học tốt!
x^3 = 27
x^3 = 3^3
Vậy x = 3
Đề bài 2 hình như sai bạn ạ
( x - 2 )^2 = 16
( x- 2 )^2 = 4^2
x - 2 = 4
x = 4 + 2
x = 6
Vậy x = 6
a) \(3\left(2x-5\right)+125=134\)
\(\Leftrightarrow3\left(2x-5\right)=9\)
\(\Leftrightarrow2x-5=3\)
\(\Leftrightarrow2x=8\Leftrightarrow x=4\)
b) \(\left(2x+5\right)+\left(2x+3\right)+\left(2x+1\right)=27\)
\(\Leftrightarrow6x+9=27\)
\(\Leftrightarrow6x=18\Leftrightarrow x=3\)
d) \(27\left(x-27\right)-27=0\)
\(\Leftrightarrow27\left(x-27\right)=27\)
\(\Leftrightarrow x-27=1\Leftrightarrow x=28\)
a) Ta có: ( 2 x + 1 ) 3 = 3 3 nên 2x + 1 = 3. Do đó x = 1.
b) Ta có: ( 2 x - 1 ) 3 = 5 3 nên 2x - 1 = 5. Do đó x = 3.
a) \(x+546=46\\ x=46-546\\ x=-500\)
b) \(2x-19\times3=27\\ 2x-57=27\\ 2x=27+57\\ 2x=84\\ x=84:2\\ x=42\)
c) \(x+12=23+3\times3^4\\ x+12=23+3\times81\\ x=23+243-12\\ x=254\)
d) \(x-12=3-3\times2^4\\ x-12=3-3\times16\\ x=3-48+12\\ x=-33\)
e) \(\left(27-x\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}27-x=0\\x+9=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=27\\x=-9\end{matrix}\right.\)
f) \(\left(-x\right)\left(x-43\right)=0\\ \Rightarrow\left[{}\begin{matrix}-x=0\\x-43=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=43\end{matrix}\right.\)
\(\frac{2x-1}{3}=\frac{27}{2x-1}\)
\(\Rightarrow\left(2x-1\right)\left(2x-1\right)=27\cdot3\)
\(\Rightarrow\left(2x-1\right)^2=81\)
\(\Rightarrow\left(2x-1\right)^2=\left(\pm9\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
ko bt đề đúng ý bn chưa ?
\(\frac{2x-1}{3}=\frac{27}{2x-1}\)
\(\Leftrightarrow\left(2x-1\right)^2=27.3=81\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=9\\2x-1=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=10\\2x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=-4\end{cases}}}\)
Lời giải:
a. $121-3(x-5)=6$
$3(x-5)=121-6=115$
$x-5=115:3=\frac{115}{3}$
$x=\frac{115}{3}+5=\frac{130}{3}$
b.
$2x-138=2^3.3^2=72$
$2x=72+138=210$
$x=210:2=105$
c.
$x-3\vdots 7$
$\Rightarrow x-3\in\left\{0;7;14;21;28;35;42;49; 56;...\right\}$
Mà $10< x< 50$ nên $x\in\left\{14;21;28;35;42;49\right\}$
d.
$27\vdots x+1$
$\Rightarrow x+1\in\left\{\pm 1; \pm 3; \pm 9; \pm 27\right\}$
$\Rightarrow x\in\left\{0; -2; -4; 2; 8; -10; 26; -28\right\}$
a ) 121-3.(x - 5 ) = 6
3.(x-5) = 121 -6
3. (x-5)=115
x-5 = 115:3
x-5=35
x=35+5
x = 40
b) 2x - 138 = 2'3. 3'2
2x -138=8.9
2x-138=72
2x=72+138
2x=210
x=210:2
x=105
c) theo bài ra : x-3 ∈ B(7)
ta có B(7)=(0,7,14,21,28,35,49,56,...)
=) x-3 ∈ ( 0,7,14,21,28,35,49,56,...)
=) x ∈( 3 , 10,17,24,31,38,42,58,..)
mà 10 <x<50 nên x ∈ ( 17 , 24 ,31,38,42 )
vậy x ∈(17,24,31,38,42)
Giải :
3^2x = 27. 3^x
=> 3^2x = 3^3 . 3^3x
=> 3^2x = 3 ^ (3+x)
=> 2x = 3+x
=> x= 3
3^2x = 27 . 3^x
3^2x = 3^3 . 3^x
3^2x = 3^3+x
2x = 3+x
2x - x = 3
x = 3