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vì x/2 =y/3=z/4 nên x2/4 = y2/ 9 = 2z2/32
áp dụng .............................
=> x2/4 = y2 /9 = 2z2 /32 = x2-y2+2z2 / 4 -9 +32 = 108 / 27 =4
=> x2 = 16 => x = 4
y2 =36 => y = 6
2z2 = 128 => z =8
đặt x/2 = y/3 = z/4 =k ( k khác 0 )
=> x = 2k
y=3k
z =4k
=> xyz = 2k3k4k = 24k = -480 => k= -20
=> x=-40
y=-60
z=-80
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\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow2^x+2^x.2+2^x.2^2+2^x.2^3=480\)
\(\Rightarrow2^x.\left(1+2+2^2+2^3\right)=480\)
\(\Rightarrow2^x.15=480\)
\(\Rightarrow2^x=480:15\)
\(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
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2x + 2x + 1 + 2x + 2 + 2x + 3 = 480
2x + 2x . 2 + 2x . 22 + 2x . 23 = 480
2x . ( 1+ 2 + 22 + 23) = 480
2x . 15 = 480
2x = 32
2x = 25
=> x = 5
Vậy x = 5
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a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110
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a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5
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\(2^{x+4}-2^x=480\)
\(\Rightarrow2^x\cdot2^4-2^x=480\)
\(\Rightarrow2^x\left(2^4-1\right)=480\)
\(\Rightarrow2^x\cdot15=480\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)