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3+2x-1=24-[16-3]
Ta co:3+2x-1=24-13
3+2x-1=11
2x-1=8
2x-1=23
x-1=3
x=4
Vậy x=4
1) 2x+1+2x=2x(2+1)=24
=>2x=8 =>x=3
2) 4x+1-3.4x=4x(4-3)=>4x=16=>x=2
3) x2-x=x(x-1)=0
=>\(\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
a,Ta có: \(\left(x+2\right)^4=\left(x+2\right)^6\)
\(\left(x+2\right)^4-\left(x+2\right)^6=0\)
\(\left(x+2\right)^4\text{[}1-\left(x+2\right)^2\text{]=0}\)
\(\Rightarrow\orbr{\begin{cases}\left(x+2\right)^4=0\\1-\left(x+2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)
\(\Rightarrow3+2^{x-1}=24-16+\left(4-1\right)\)
\(\Rightarrow3+2^{x-1}=8+3\)
\(\Rightarrow2^{x-1}=2^3\)
\(\Rightarrow x=4\)
\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)
\(\Rightarrow3+2^{x-1}=24-\left[16-\left(4-1\right)\right]\)
\(\Rightarrow3+2^{x-1}=24-\left[16-3\right]\)
\(\Rightarrow3+2^{x-1}=24-13\)
\(\Rightarrow3+2^{x-1}=11\)
\(\Rightarrow2^{x-1}=11-3\)
\(\Rightarrow2^{x-1}=8\)
\(\Rightarrow2^{x-1}=2^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=3+1\)
\(\Rightarrow x=4\)
Vậy x = 4
_Chúc bạn học tốt_
\(3+2^{x-1}=24-\left\{4^2-\left(2^2-1\right)\right\}\)
\(\Rightarrow3+2^{x-1}=24-\left\{16-\left(4-1\right)\right\}\)
\(\Rightarrow3+2^{x-1}=24-\left\{16-3\right\}\)
\(\Rightarrow3+2^{x-1}=24-13\)
\(\Rightarrow3+2^{x-1}=11\)
\(\Rightarrow2^{x-1}=11-3\)
\(\Rightarrow2^{x-1}=8\)
\(\Rightarrow2^{x-1}=2^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=3+1\)
\(\Rightarrow x=4\)
Vậy x = 4
_Chúc bạn học tốt_
a)(2x-2)3=8
(2x-2)3=(\(\pm\)2)3
Vậy 2x-2=2 hoặc 2x-2=(-2)
2x =2+2=4 2x =2+(-2)=0
x =4:2=2 x 0:2=0
Câu b mik ko biết sorry nha!
Hok tốt!
\(2^x+2^{x+1}=24\\ \Rightarrow2^x+2^x\cdot2=24\\ \Rightarrow2^x\left(2+1\right)=24\\ \Rightarrow2^x\cdot3=24\\ \Rightarrow2^x=24:3\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)