4/2x+3/=36
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\(24x-4\left(2x-\frac{3}{4}\right)-4\left(3+\frac{2x}{2}\right)=36-3\left(x-\frac{3}{2}\right)-3\left(3-\frac{2x}{3}\right)\)
Đề như này đúng không bạn
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[\left(2x-4\right)+9\right]^3=3^6\)
\(\Leftrightarrow2x-4=0\)
hay x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
2x - 36 = 46 : 43
2x - 36 = 64
2x = 64 + 36
2x = 100
x = 100 : 2
x = 50
5x+ 11= 36
5x = 36 - 11
5x = 25
=> x = 2
k mik nha làm ơn
a)\(2x-36=4^6:4^3\Rightarrow2x-36=4^3\Rightarrow2x-36=64\Rightarrow2x=64+36\Rightarrow2x=100\Rightarrow x=50\)
b)\(5^x+11=36\Rightarrow5^x=36-11\Rightarrow5^x=25\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>4x-15-2x+3=36
=>2x-12=36
=>2x=48
hay x=24
b: =>2x-x=5+4
hay x=9
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
( 2x - 1 )( 4 + 3 ) = - 36
=> 8x + 6x - 4 - 3 = -36
=> 14x - 7 = -36
=> 14x = -29
=> x = -29/14
Vậy x = -29/14
![](https://rs.olm.vn/images/avt/0.png?1311)
3) \(...\Rightarrow2^x\left(2^3+1\right)=36\)
\(\Rightarrow2^x.9=36\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\Rightarrow x=2\)
4) \(...\Rightarrow4^{x+1}-4^x=12\)
\(\Rightarrow4^x\left(4-1\right)=12\)
\(\Rightarrow4^x.3=12\)
\(\Rightarrow4^x=4=4^1\Rightarrow x=1\)
5) \(...\Rightarrow5^{x+1}\left(5^2-1\right)=3000\)
\(\Rightarrow5^{x+1}.24=3000\)
\(\Rightarrow5^{x+1}=125\)
\(\Rightarrow5^{x+1}=5^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
6) Bạn xem lại đề
a. \(2^x.2^3+2^x=36\)
\(2^x\left(2^3+1\right)=36\)
\(2^x.9=36\)
\(2^x=4\Rightarrow x=2\)
b. \(4^x.4^1-\left(2^2\right)^x=12\)
\(4^x.4-4^x=12\)
\(4^x\left(4-1\right)=12\)
\(4^x.3=12\)
\(4^x=4\)
x = 1
c. \(5^x.5^3-5^x.5^1=3000\)
\(5^x\left(5^3-5^1\right)=3000\)
\(5^x.120=3000\)
\(5^x=25\)
x = 2
d. \(4^{x+1}=2^{2x}\)
\(4^x.4=\left(2^2\right)^x\)
\(4^x.4=4^x\)
Có vẻ như câu 4 này để bài thiếu
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x^3+4x^2-6x^2-24x+9x+36\right):\left(x+4\right)\\ =\left[x^2\left(x+4\right)-6x\left(x+4\right)+9\left(x+4\right)\right]:\left(x+4\right)\\ =x^2-6x+9\)
4/2x+3/ = 36
/2x+3/ = 36 : 4
/2x+3/ = 9
\(\Rightarrow\left[{}\begin{matrix}2x+3=9\\2x+3=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=9-3\\2x=\left(-9\right)-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-12\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-6\end{matrix}\right.\)
Vậy x \(\in\) { 3;-6 }
4/2x+3/ = 36
/2x+3/ = 36 : 4
/2x+3/ = 9
⇒[2x+3=92x+3=−9⇒[2x+3=92x+3=−9
⇒[2x=9−32x=(−9)−3⇒[2x=9−32x=(−9)−3
⇒[2x=62x=−12⇒[2x=62x=−12
⇒[x=6:2x=(−12):2⇒[x=6:2x=(−12):2
⇒[x=3x=−6⇒[x=3x=−6
Vậy x ∈∈ { 3;-6 }