Tìm x cho biết: \(12^2< x< 200\)
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a) Ta thấy:
\(\left(x-3\right)^2\ge0\)
\(\left(y+2\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Để \(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\begin{cases}\left(x-3\right)^2=0\\\left(y+3\right)^2=0\end{cases}\)
\(\Rightarrow\begin{cases}x-3=0\\y+3=0\end{cases}\)
\(\Rightarrow\begin{cases}x=3\\y=-3\end{cases}\)
Vậy \(\begin{cases}x=3\\y=-3\end{cases}\)
c) Ta thấy:
\(\left(x-12+y\right)^{200}\ge0\)
\(\left(x-4-y\right)^{200}\ge0\)
\(\Rightarrow\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}\ge0\)
Để \(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
\(\Rightarrow\begin{cases}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{cases}\)
\(\Rightarrow\begin{cases}x-12+y=0\\x-4-y=0\end{cases}\)
\(\Rightarrow\begin{cases}x+y=12\\x-y=4\end{cases}\)
\(\Rightarrow\begin{cases}x=\left(12+4\right):2\\y=\left(12-4\right):2\end{cases}\)
\(\Rightarrow\begin{cases}x=8\\y=4\end{cases}\)
Vậy \(\begin{cases}x=8\\y=4\end{cases}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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(-12)2.x = 56 + 10.13.x
144.x = 56 + 130.x
144x – 130 x = 56
14x = 56
x = 56: 14
x = 4
Vậy x = 4
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Bài 1 :
\(8^7-2^{18}\)
\(=\left(2^3\right)^7-2^{18}\)
\(=2^{21}-2^{18}\)
\(=2^{18}\left(2^3-1\right)\)
\(=2^{18}\cdot7\)
\(=2^{17}\cdot2\cdot7\)
\(=2^{17}\cdot14⋮14\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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A=1/1.2+1/2.3+...1/x =49/50
A=1-1/2+1/2-1/3+...+1/x-1-1/x=49/50
A=1-1/x=49/50
A=50/50-1=x=49/50
x=1/50
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x-1\right)}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x-1}-\frac{1}{x}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{x}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{50}\)
\(\Rightarrow x=50\)
122 < x < 200
144 < x < 200
=> x \(\in\){ 145 ; 146 ; ... ; 199 }
Vậy x \(\in\){ 145 ; 146 ; ... ; 199 }
\(12^2< x< 200\)
\(12^2< x< \sqrt{200}\)
\(12^2< x< \left(14,142...\right)^2\)
\(\Rightarrow x=13^2\)
vậy \(x=13\)