Tìm x biết 2x=16
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`@` `\text {Ans}`
`\downarrow`
`2^x = 16`
`=> 2^x = 2^4`
`=> x = 4`
Vậy, `x = 4.`
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`2^x*16 = 1024`
`=> 2^x =`\(2^{10}\div2^4\)
`=> 2^x = 2^6`
`=> x = 6`
Vậy, `x = 6`
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`2^x - 26 = 6`
`=> 2^x = 6 + 26`
`=> 2^x = 32`
`=> 2^x = 2^5`
`=> x = 5`
Vậy, `x = 5`
`3^x*3 = 243`
`=> 3^x * 3 = 3^5`
`=> 3^x = 3^5 \div 3`
`=> 3^x = 3^4`
`=> x = 4`
Vậy, `x = 4.`
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Ta có: \(2\left(x-1\right)-3\left(2x+2\right)-4\left(2x+3\right)=16\)
\(\Rightarrow2x-2-6x-6-8x-12=16\)
\(\Rightarrow2x-6x-8x=16+2+6+12\)
\(\Rightarrow-12x=36\)
\(\Rightarrow x=-3\)
Vậy x = -3
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a) \(\left(2x-1\right)^4=16\)
\(\)TH1: \(\left(2x-1\right)^4=2^4\)
\(=>2x-1=2\)
\(2x=2+1\)
\(2x=3\)
\(x=\dfrac{3}{2}\)
TH2: \(\left(2x-1\right)^4=\left(-2\right)^4\)
\(=>2x-1=-2\)
\(2x=-2+1\)
\(2x=-1\)
\(x=\dfrac{-1}{2}\)
Vậy x = \(\dfrac{3}{2}\) hoặc x = \(\dfrac{-1}{2}\)
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b) \(\left(2x+1\right)^3=125\) ( mình nghĩ đề bài đúng là vầy )
\(\left(2x+1\right)^3=5^3\)
\(=>2x+1=5\)
\(2x=5-1\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
Vậy x = \(2\)
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Ta có : \(\left(2x+2016\right)^3=\left(x+2000\right)^3+\left(x+16\right)^3\)
=> \(\left(2x+2016\right)^3-\left(x+2000\right)^3-\left(x+16\right)^3=0\)(*)
Gọi \(a=x+2000 ; b=x+16\)
=> ;\(a+b=2x+2016\)
Từ (*) suy ra : \(\left(a+b\right)^3-a^3-b^3=0\)
=> \(3ab\left(a+b\right)=0\)
+) \(a=0\) => \(x+2000=0\) => \(x=-2000\)
+) \(b=0\) => \(x+16=0\) => \(x=-16\)
+) \(a+b=0\)=> \(2x+2016=0\) => \(x=-1008\)
Vậy \(x\in\left\{-2000;-1008;-16\right\}\)
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|5(2x+3)| + |2(2x+3)| + |2x+3| = 16
|5| * |2x+3| + |2| * |2x+3| + |2x+3| = 16
8* |2x+3| = 16
|2x+3|= 2
\(\Rightarrow\orbr{\begin{cases}2x+3=2\\2x+3=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}}\)
vậy x= -1/2 hoặc x= -5/2
Vì |5(2x+3)|, |2(2x+3)| và |2x+3| luôn luôn là số tự nhiên
=> 5(2x+3)+2(2x+3)+2x+3 = 16
<=> (2x+3).(5+2+1) = 16
<=> 2x+3 = 16:8 = 2
<=> 2x = 2-3 = -1
<=> x = -1:2 = -1/2
2x=16
suy ra 2x=2 mũ 4
suy ra x=4
thông cảm nhé
2x=16
2x=24
x=4
Vay x=4