2x+3.(2x-4)=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
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
a: =>4x-15-2x+3=36
=>2x-12=36
=>2x=48
hay x=24
b: =>2x-x=5+4
hay x=9
\(\left[\left(2x-4\right)+9\right]^3=3^6\)
\(\Leftrightarrow2x-4=0\)
hay x=2
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
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 }
\(\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=4\\ \Rightarrow x=2.\\ b,\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}.\\ c,6.3^x-2.3^x=36\\ \Rightarrow3^x.\left(6-2\right)=36\\ \Rightarrow3^x.4=36\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2.\\ d,2^{x+1}-2^x=32\\ \Rightarrow2^x.\left(2-1\right)=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5.\)
2x+6x-12=36
8x=48
x=6
x = 6 nhé Nguyễn Diệp Anh