tìm x biết
\(4^x-12.2^x+32=0\)
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D
29 tháng 12 2019
\(4^x-12.2^x+32=0\Leftrightarrow\left(2^x\right)^2-2.6.2^x+6^2-4=0\Leftrightarrow\left(2^x-6\right)^2-2^2=0\)
\(\Leftrightarrow\left(2^x-6-2\right)\left(2^x-6+2\right)=0\Leftrightarrow\left(2^x-8\right)\left(2^x-4\right)=0\Leftrightarrow\orbr{\begin{cases}2^x-8=0\\2^x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2^x=8\\2^x=4\end{cases}\Leftrightarrow\orbr{\begin{cases}2^x=2^3\\2^x=2^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
Vậy \(S=\left\{2;3\right\}\)
3 tháng 1 2018
\(4^x-12.2^x+32=0\)
⇒ \(2^x.2^x-4.2^x-8.2^x+4.8=0\)
⇒ \(2^x\left(2^x-4\right)-8\left(2^x-4\right)=0\)
⇒ \(\left(2^x-4\right)\left(2^x-8\right)=0\)
⇒ \(\left[{}\begin{matrix}2^x-4=0\\2^x-8=0\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}2^x=2^2\\2^x=2^3\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
4 tháng 2 2020
Ta có : \(4^x+12.2^x+32=0\)
<=> \(\left(2^x\right)^2+2.6.2^x+36-4=0\)
<=> \(\left(2^x+6\right)^2-4=0\)
<=> \(\left(2^x+6+2\right)\left(2^x+6-2\right)=0\)
<=> \(\left(2^x+8\right)\left(2^x+4\right)=0\)
<=> \(\left[{}\begin{matrix}2^x+8=0\\2^x+4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}2^x=-8\\2^x=-8\end{matrix}\right.\) ( Vô lý )
Vậy phương trình vô nghiệm .
2 tháng 1 2023
=>(2^x)^2-12*2^x+32=0
=>(2^x-4)(2^x-8)=0
=>x=3 hoặc x=2
ND
14 tháng 4 2018
a)
\(x^2-4x+4=25\)
\(\Leftrightarrow x^2-4x-21=0\)
\(\Leftrightarrow x^2+3x-7x-21=0\)
\(\Leftrightarrow x\left(x+3\right)-7\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b)
\(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)
\(\Leftrightarrow\dfrac{x-17}{1990}-1+\dfrac{x-21}{1986}-1+\dfrac{x+1}{1004}-2=4-1-1-2\)
\(\Leftrightarrow\dfrac{x-17-1990}{1990}+\dfrac{x-21-1986}{1986}+\dfrac{x+1-2008}{1004}=0\)
\(\Leftrightarrow\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\right)=0\)
\(\Leftrightarrow x-2007=0\) ( Vì: \(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\ne0\))
\(\Leftrightarrow x=2007\)
ND
14 tháng 4 2018
c.
\(4^x-12.2^x+32=0\)
\(\Leftrightarrow\left(2^x\right)^2-12.2^x+36-4=0\)
\(\Leftrightarrow2^x-2.2^x.6+6^2-2^2=0\)
\(\Leftrightarrow\left(2^x-6\right)^2-2^2=0\)
\(\Leftrightarrow\left(2^x-6-2\right)\left(2^x-6+2\right)=0\)
\(\Leftrightarrow\left(2^x-8\right)\left(2^x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x-8=0\\2^x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x=8\\2^x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
đặt: \(\left\{{}\begin{matrix}2^x=t\\t>0\end{matrix}\right.\)
\(t^2-12t+32=0\Leftrightarrow t^2-2.6t+36=4\)
\(\left(t-6\right)^2=2^2\Rightarrow\left[{}\begin{matrix}t=8\\t=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Ta có : 4x = (2x)2 .
=> 4x - 12.2x + 32 = 0 <=> (2x)2 - 12.2x + 36 - 4 = 0
<=> (2x - 6 )2 - 4 = 0
<=> (2x - 6 - 2 ).( 2x - 6 + 2 ) = 0
<=> ( 2x - 8 ).( 2x - 4 ) = 0 .
=> \(\left[{}\begin{matrix}2^x=8\\2^x=4\end{matrix}\right.\) => \(\left[{}\begin{matrix}2^x=2^3\\2^x=2^2\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)