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\(2^{2x+1}+4^x=48\)
\(\Leftrightarrow2^{2x+1}+2^{2x}=48\)
\(\Leftrightarrow2^{2x}\left(2+1\right)=48\)
\(\Leftrightarrow2^{2x}=16\)'
\(\Leftrightarrow2^{2x}=2^4\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
\(\left(x^2-25\right)\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-25=0\\x^2+1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x^2=25\\x^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=5\\x=-5\end{cases}}\\\orbr{\begin{cases}x=1\\x=-1\end{cases}}\end{cases}}\)
\(3\left(x^2-4\right)-\left(2x^2-1\right)=38\)
\(3x^2-12-2x^2+1=38\)
\(\left(3x^2-2x^2\right)-12+1=38\)
\(1x^2-11=38\)
\(x^2=49\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
Thiếu :(
\(x^2=25\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
\(x^2=1\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
123-5.(x+4)=38
5(x+4)=123-38
5(x+4)=85
x+4=85:5
x+4=17
x=17-4
x=13
10+2x=2.(32-1)
10+2x=2.(9-1)
10+2x=2.8
2x=16-10
2x=6
x=6:2
x=3
2.x-138=23.32
2.x-138=72
2.x=72+138
2.x=210
x=210:2
x=105
[102+(28-2x)]:20-5=1
(102+28-2x):20=6
130-2x=120
2x=130-120
2x=10
x=10:2
x=5
231-(x-6)=1339:13
231-(x-6)=103
x-6=231-103
x-6=128
x=128+6
x=134
123-5.(x+4)=38
5.(x+4)=123-38
5.(x+4)=85
(x+4)=85/5
x+4=17
x=17-4
x=13
a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)
- 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50
2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51
7\(x\) = 67
\(x\) = 67 : 7
\(x\) = \(\dfrac{67}{7}\)
Vậy \(x\) = \(\dfrac{67}{7}\)
b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)
17\(x\) - 48\(x\) - 111 = 2\(x\) - 4\(x\) + 43
- 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43
- \(x\) x (31 + 2 - 4) = 154
- \(x\) x (33 - 4) = 154
- \(x\) x 29 = 154
- \(x\) = 154 : (-29)
\(x\) = - \(\dfrac{154}{29}\)
Vậy \(x=-\dfrac{154}{29}\)
6x . 6 = 2016
6x = 2016 : 6
6x = 336
=> x \(\in\varnothing\)
42x+3 : 4 = 256
42x+3 = 256 x 4
42x+3 = 1024
42x+3 = 45
2x + 3 = 5
2x = 5 - 3
2x = 2
x = 2 : 2
x = 1
[ x - 2 ]2 = 16
[ x - 2 ]2 = 42
x - 2 = 4
x = 4 + 2
x = 6
[ 2x - 1 ]3 = 27
[ 2x - 1 ]3 = 33
2x - 1 = 3
2x = 3 + 1
2x = 4
x = 4 : 2
x = 2
[ 2x - 1 ]100 = [ 2x - 1 ]100
=> x \(\in N\)
\(\left(2x-4\right)^{38}=\left(2x-4\right)^{48}\)
\(\Rightarrow\left(2x-4\right)^{38}-\left(2x-4\right)^{48}=0\)
\(\Rightarrow\left(2x-4\right)^{38}\left[1-\left(2x-4\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-4\right)^{38}=0\\1-\left(2x-4\right)^{10}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(2x-4\right)^{10}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x-4=1\\2x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\\2x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{2;\dfrac{5}{2};\dfrac{3}{2}\right\}.\)
#\(Toru\)
thank you