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\(\Rightarrow\left(2x-1\right)^{100}-\left(2x-1\right)^{80}=0\)
Đặt 2x - 1 = t ta có :
\(t^{100}-t^{80}=0\Rightarrow t^{80}\left(t^{20}-1\right)=0\)
=> \(t^{80}=0\) hoặc \(t^{20}-1=0\)
=> t = 0 hoặc \(t^{20}=1=\left(-1\right)^2\)
=> t= 0 hoặc t = 1 hoặc t = -1
(+) t = 0 => 2x - 1 = 0 => x = 1/2
(+) t = 1 => 2x - 1 = 1 => x = 1
(+) t= - 1 => 2x - 1 = -1 => x = 0
tick đúng nha
=> (2x -1)100 - (2x -1)80 = 0
=> (2x -1)80. [(2x-1)20 - 1] = 0
=> (2x -1)80 = 0 hoặc (2x -1)20 = 1
+) (2x -1)80 = 0 => 2x - 1 = 0 => x = 1/2
+) (2x -1)20 = 1 => 2x - 1 = 1 hoặc 2x - 1 = -1
=> 2x = 2 hoặc 2x = 0 => x = 1 hoặc x = 0
Vậy....
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a) \(3^{x+2}\cdot5^{y-3}=45^x\)
\(\Rightarrow3^{x+2}\cdot5^{y-3}=\left(3^2\right)^x\cdot5^x\)
\(\Rightarrow3^{x+2}\cdot5^{y-3}=3^{2x}\cdot5^x\)
\(\Rightarrow\left\{{}\begin{matrix}3^{x+2}=3^{2x}\\5^{y-3}=5^x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=2x\\y-3=x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y-3=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)
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1) \(5^{x+1}-5^x=20\Leftrightarrow5^x\left(5-1\right)=20\Leftrightarrow5^x=5\Leftrightarrow x=1\)
2) \(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\Leftrightarrow2^x=32\Leftrightarrow x=5\)
3) \(4^{2x+1}+4^{2x}=80\Leftrightarrow4^{2x}\left(4+1\right)=80\Leftrightarrow16^x=16\Leftrightarrow x=1\)
4) \(3^{2x+2}+3^{2x+1}=108\Leftrightarrow3^{2x}\left(3^2+3\right)=108\Leftrightarrow9^x=9\Leftrightarrow x=1\)
5) \(7^{x+3}-7^{x+1}=16464\Leftrightarrow7^x\left(7^3-7\right)=16464\Leftrightarrow7^x=49\Leftrightarrow x=2\)
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làm bài này nữa rùi học bài
1. 4/2 = -4x2
x2 = -1/2 vô nghiệm
2. 2x(1+4) = 80
2x = 16 = 24
x = 4
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\(\left|x-1\right|+\left|x+5\right|=\left|x-1\right|+\left|-x-5\right|\)
\(\Rightarrow\left|x-1\right|+\left|x+5\right|\ge\left|x-1-x-5\right|\)
\(\Rightarrow\left|x-1\right|+\left|x+5\right|\ge\left|-6\right|=6\)
dấu "=" xảy ra khi \(\left(x-1\right).\left(x+5\right)\ge0\)
\(\Rightarrow-5\le x\le1\)
Vậy x={-5,-4,-3,-2,-1,0,1}
b) \(\hept{\begin{cases}\left(2x-y+3\right)^4\ge0\\\left|y+2\right|\ge0\end{cases}}\)
mà \(\left(2x-y+3\right)^4+\left|y+2\right|=0\)
dấu "=" xảy ra khi \(\hept{\begin{cases}\left(2x-y+3\right)^4=0\\\left|y+2\right|=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{5}{2}\\y=-2\end{cases}}\)
vậy \(x=-\frac{5}{2},y=-2\)
dấu "=" xảy ra khi
Vậy x={-5,-4,-3,-2,-1,0,1}
b)
mà
dấu "=" xảy ra khi
vậy
2^x+2^x+2=80
2^x(1+2^2)=80
2^x.5=80
2^x=16
2^x=2^4=>x=4