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`4^(x+4)-4^x =4080`
`=>4^x *4^4 -4^x =4080`
`=>4^x (256-1)=4080`
`=>4^x *255=4080`
`=>4^x =16`
`=>4^x =4^2`
`=>x=2`
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\(=>4xx-4x+1=0\)
\(4x\left(x-1\right)+1=0\)
\(=>\left[{}\begin{matrix}4x=0\\\left(x-1\right)+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x-1=-1\end{matrix}\right.\left[{}\begin{matrix}x=0\\x=0\end{matrix}\right.\)
vậy x=0
\(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
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a.
\(4x-8⋮2x+3\Rightarrow4x+6-14⋮2x+3\)
\(\Rightarrow2\left(2x+3\right)-14⋮2x+3\)
\(\Rightarrow14⋮2x+3\)
\(\Rightarrow2x+3=Ư\left(14\right)\)
Do \(2x+3\) luôn lẻ khi x nguyên nên ta chỉ cần xét các ước lẻ của 14
\(\Rightarrow2x+3=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x=\left\{-5;-2;-1;2\right\}\)
b.
\(2xy+4x-3y=17\)
\(\Leftrightarrow2xy-3y+4x-6=17-6\)
\(\Leftrightarrow y\left(2x-3\right)+2\left(2x-3\right)=11\)
\(\Leftrightarrow\left(2x-3\right)\left(y+2\right)=11\)
Bảng giá trị:
2x-3 | -11 | -1 | 1 | 11 |
y+2 | -1 | -11 | 11 | 1 |
x | -4 | 1 | 2 | 7 |
y | -3 | -13 | 9 | -1 |
Vậy \(\left(x;y\right)=\left(-4;-3\right);\left(1;-13\right);\left(2;9\right);\left(7;-1\right)\)
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\(\left(4x-15\right)^{2016}=\left(4x-15\right)^{2015}\\ \Leftrightarrow\left[{}\begin{matrix}4x-15=0\\4x-15=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x=15\\4x=16\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{15}{4};4\right\}\)
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a) \(x^3+4x=0\)
\(\Rightarrow x\left(x^2+4\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2+4=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x^2=-4\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x\in\phi\end{array}\right.\)
Vậy: \(x=0\)
b) \(2\left(5-x\right)=4x-3\)
\(\Rightarrow10-2x=4x-3\)
\(\Rightarrow10+3=4x+2x\)
\(\Rightarrow13=6x\)
\(\Rightarrow x=\frac{13}{6}\)
x3+ 4x=0
<=> x(x2+4)=0
=> x=0 hoặc x2+4=0
Mà: x2+4 >4
=>x=0
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