( 2x -4)38= ( 2x-4)48
K
Khách
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H
2
2 tháng 9 2015
5.x - 38:19 = 13
=> 5x - 2 = 13
=> 5x = 15
=> x = 3
84-4(2x+1) = 48
=> 4(2x+1) = 36
=> 2x + 1 = 9
=> 2x = 10
x = 5
BE
0
30 tháng 8 2021
a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)
\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)
\(\Leftrightarrow9x^2=16\)
\(\Leftrightarrow x^2=\dfrac{16}{9}\)
hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)
K
1
24 tháng 3 2020
(2x2 + 1)(x-3)=0
\(\Rightarrow\orbr{\begin{cases}2x^2+1=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x^2=-1\Rightarrow x^2=-\frac{1}{2}\left(vl\right)\\x=3\end{cases}}\)
Vậy x=3
48-(15-x)5=48
(15-x)5=48-48
(15-x)5=0
=> 15-x =0
x =15-0
x =15
Vậy x=15
CT
2
KV
18 tháng 9 2023
2x - 3 = 17
2x = 17 + 3
2x = 20
x = 20 : 2
x = 10
---------------
4² - 3x = 27.3 - 77
16 - 3x = 81 - 77
16 - 3x = 4
3x = 16 - 4
3x = 12
x = 12 : 3
x = 4
----------------
48 + 2x = 3.2³
48 + 2x = 3.8
48 + 2x = 24
2x = 24 - 48
2x = -24
x = -24 : 2
x = -12
----------------
(13x - 12²) : 5 = 5
13x - 144 = 5.5
13x - 144 = 25
13x = 25 + 144
13x = 169
x = 169 : 13
x = 13
VN
Võ Ngọc Phương
CTVHS
VIP
18 tháng 9 2023
\(2x-3=17\)
\(2x=17+3\)
\(2x=20\)
\(x=20:2\)
\(x=10\)
____
\(4^2-3x=27.3-77\)
\(16-3x=81-77\)
\(16-3x=4\)
\(3x=16-4\)
\(3x=12\)
\(x=12:3\)
\(x=4\)
_____
\(48+2x=3.2^3\)
\(48+2x=3.8=24\)
\(2x=24-48\)
\(2x=\left(-24\right)\)
\(x=\left(-24\right):2\)
\(x=\left(-12\right)\)
____
\(\left(13x-12^2\right):5=5\)
\(\left(13x-144\right):5=5\)
\(13x-144=5.5\)
\(13x-144=25\)
\(13x=25+144\)
\(13x=169\)
\(x=169:13\)
\(x=13\)
NT
0
17 tháng 1 2021
a, \(2x+3⋮2x-1\)
\(2x-1+4⋮2x-1\)
\(4⋮2x+1\)hay \(2x+1\inƯ\left(4\right)=\left\{1;2;4\right\}\)
| 2x + 1 | 1 | 2 | 4 |
| 2x | 0 | 1 | 3 |
| x | 0 | 1/2 | 3/2 |
c, \(\left(x+5\right)\left(y-3\right)=15\Leftrightarrow x+5;y-3\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)
| x + 5 | 1 | 3 | 5 | 15 |
| y - 3 | 15 | 5 | 3 | 1 |
| x | -4 | -2 | 0 | 10 |
| y | 18 | 8 | 6 | 4 |
\(\left(2x-4\right)^{38}=\left(2x-4\right)^{48}\)
\(\Rightarrow\left(2x-4\right)^{38}-\left(2x-4\right)^{48}=0\)
\(\Rightarrow\left(2x-4\right)^{38}\left[1-\left(2x-4\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-4\right)^{38}=0\\1-\left(2x-4\right)^{10}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(2x-4\right)^{10}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x-4=1\\2x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\\2x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{2;\dfrac{5}{2};\dfrac{3}{2}\right\}.\)
#\(Toru\)
thank you