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Ta có : (6 - x)2014 = (6 - x)2015
=> (6 - x)2014 - (6 - x)2015 = 0
<=> (6 - x)2014(1 - 6 - x) = 0
<=> \(\orbr{\begin{cases}\left(6-x\right)^{2014}=0\\1-6-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}6-x=0\\-5-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=6\\x=-5\end{cases}}\)
sory bạn trừng hợp hai mk nhầm :
1 - (6 - x) = 0
=> 1 - 6 + x = 0
=> -5 + x = 0
=> x = 5
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\(2^x\)+\(2^x\).2+\(2^x\).\(2^2\)=28
\(2^x\)(1+2+4)=28
\(2^x\).7=28
\(2^x\)=4
\(2^2\)=4
x=4
a,
b, \(2^x+2^{x+1}+2^{x+2}=28\)
\(\Leftrightarrow2^x+2^x.2+2^x.2^2=28\)
\(\Leftrightarrow2^x\left(1+2+4\right)=28\)
\(\Leftrightarrow2^x.7=28\)
\(\Leftrightarrow2^x=4\)
\(\Leftrightarrow2^x=2^2\)
=> x = 2
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a) 120 - 5 . ( x + 2 ) = 45
5 . (x + 2) = 120 - 45
5 . (x + 2) = 75
x + 2 = 75 : 5
x + 2 = 15
x = 17
b) ( 2.x - 3 )2 = 49
( 2.x - 3 )2 = 72
( 2.x - 3 ) = 7
2x = 10
x = 5
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<=> 7 + \(2^{x+1}\)=28-[16-3]
<=>7 + \(2^{x+1}\)=28-13
<=>7 + \(2^{x+1}\)=15
<=>\(2^{x+1}=8\)
<=>\(2^{x+1}=2^3\)
<=>x+1=3
<=>x=2
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a,3(x+1)+2x+5=28
\(\Rightarrow\)3x+3+2x+5=28
\(\Rightarrow\)5x+8=28
\(\Rightarrow\)5x=20
\(\Rightarrow\)x=4
b,3x-1=92
\(\Rightarrow\)3x-1=(32)2
\(\Rightarrow\)3x-1=34
\(\Rightarrow\)x-1=4
\(\Rightarrow\)x=5
c,3x+2+2.3x=99
\(\Rightarrow\)3x.9+2.3x=99
\(\Rightarrow\)11.3x=99
\(\Rightarrow\)3x=9
\(\Rightarrow\)3x=32
\(\Rightarrow\)x=2
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a ) \(3\left(x+1\right)+2x+5=28\)
=> \(3x+3+2x+5=28\)
=> \(5x+8=28\)
=> \(5x=20\)
=> \(x=4\)
B ) \(3^{x-1}=9^2\)
=> \(3^{x-1}=81\)
=> \(x=5\)
\(a,
3\left(x+1\right)+2x+5=28\)
\(\Rightarrow3x+3+2x+5=28\)
\(\Rightarrow5x+8=28\)
\(\Rightarrow5x=20\)
\(\Rightarrow x=4\)
\(b,3^{x-1}=9^2\)
\(\Rightarrow3^{x-1}=81\)
\(\Rightarrow x=5\)
Sửa đề : \(2^{x-1}+2^x+2^{x+1}=32\)
\(\Leftrightarrow2^x\left(1:2.2\right)=32\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\Leftrightarrow x=5\)
Vậy x = 5
\(2^{x-1}+2^x+2^{x+1}=28\)
\(\Leftrightarrow2^x:2+2^x+2^x.2=28\)
\(\Leftrightarrow2^x.\frac{1}{2}+2^x+2^x.2=28\)
\(\Leftrightarrow2^x.\left(\frac{1}{2}+1+2\right)=28\)
\(\Leftrightarrow2^x.\frac{7}{2}=28\)
\(\Leftrightarrow2^x=8\)
\(\Leftrightarrow2^x=2^3\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)