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\(\left(x^2+2\right).\left(x^2-25\right)>0\)
\(TH1:\hept{\begin{cases}x^2+2>0\\x^2-25>0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2>-2\\x^2>25\end{cases}\Leftrightarrow}-5>x>}5\)
\(TH2:\hept{\begin{cases}x^2+2< 0\\x^2-25< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2< -2\\x^2< 25\end{cases}\Leftrightarrow-}5< x< 5}\)
a) \(\frac{x}{7}=\frac{5}{y}\left(ĐK:x>y\right)\)
\(\Leftrightarrow5.7=x.y\)
\(\Rightarrow\orbr{\begin{cases}x=7\\y=5\end{cases}}\)(Vì x > y)
b) \(\frac{2}{x}=\frac{x}{-7}\left(ĐK:x>0\right)\)
\(\Leftrightarrow2.\left(-7\right)=x.x\)
\(\Leftrightarrow\left(-14\right)=x^2\)
\(\Rightarrow x=\sqrt{14}\)
a) \(3^{x+1}.15=135\)
\(\Rightarrow3^{x+1}=9\)
\(\Rightarrow3^{x+1}=3^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
b) \(x+2x+2^2x+....+2^{2016}x=2^{2017}-1\\ \Rightarrow x\left(2+2^2+...+2^{2016}\right)=2^{2017}-1\\ \Rightarrow x\left(2^{2017}-2\right)=2^{2017}-1\)
c) \(x\left(x-1\right)+\left(x-1\right)^2=0\\ \Rightarrow x\left(x-1\right)+\left(x-1\right)\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+\left(x-1\right)\right)=0\\ \Rightarrow\left(x-1\right)\left(2x-1\right)=0\\ \Rightarrow\begin{cases}x-1=0\\2x-1=0\end{cases}\)
d) \(2^2.2^5\le2^{x-5}\le2^{10}\\ \Rightarrow2^7\le2^{x-5}\le2^{10}\)
\(2^4.32>2^x>64\\ \Leftrightarrow2^4.2^5>2^x>2^6\\ \Leftrightarrow2^9>2^x>2^6\\ \Leftrightarrow9>x>6\)
=> x=7 hoặc x=8
2⁴.32>2^x>64
=>2⁹>2^x>2⁶
=>9>x>6
=>x€{8;7}