83: \(4^x+2^x-2>0\)
=>\(\left(2^x\right)^2+2\cdot2^x-2^x-2>0\)
=>\(2^x\left(2^x+2\right)-\left(2^x+2\right)>0\)
=>\(\left(2^x+2\right)\left(2^x-1\right)>0\)
=>\(2^x-1>0\)
=>\(2^x>1\)
=>x>0
=>Chọn A
87:
\(4^x-5\cdot2^x+4< 0\)
=>\(\left(2^x\right)^2-5\cdot2^x+4< 0\)
=>\(\left(2^x\right)^2-2^x-4\cdot2^x+4< 0\)
=>\(\left(2^x-1\right)\left(2^x-4\right)< 0\)
=>\(1< 2^x< 4\)
mà \(x\in Z\)
nên x=1
=>Chọn A
90:
\(\left(\dfrac{1}{1+a^2}\right)^{2x+1}>1\)
=>\(\left(a^2+1\right)^{-2x-1}>1\)
=>\(-2x-1>0\)
=>-2x>1
=>\(x< -\dfrac{1}{2}\)
=>Chọn B
Câu 62
đk 6 - 5x > 0
\(\Leftrightarrow6-5^x=5^{1-x}=\dfrac{5}{5^x}\)
Đặt 5x = t ( t > 0 ) \(\Leftrightarrow6-t=\dfrac{5}{t}\Leftrightarrow6t-t^2=5\Leftrightarrow t^2-6t+5=0\Leftrightarrow t=1;5\)
\(\Rightarrow\left[{}\begin{matrix}5^x=1\\5^x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=log_5^1\\x=1\end{matrix}\right.\)
Câu 83
\(2^x.2^x+2^x-2>0\)
Đặt 2x = t (t>0)
\(\Leftrightarrow t^2+t-2>0\Leftrightarrow\left(t-1\right)\left(t+2\right)>0\Leftrightarrow\left[{}\begin{matrix}t>1\\t< -2\left(l\right)\end{matrix}\right.\)
=> chọn C
62.
ĐKXĐ: ...
\(log_5\left(6-5^x\right)=1-x\)
\(\Rightarrow6-5^x=5^{1-x}\)
\(\Rightarrow6-5^x=\dfrac{5}{5^x}\)
\(\Rightarrow\left(5^x\right)^2-6.5^x+5=0\)
\(\Rightarrow\left[{}\begin{matrix}5^x=1\\5^x=5\end{matrix}\right.\) \(\Rightarrow x=\left\{0;1\right\}\)
82.
\(5.4^x+2.25^x-7.10^x\le0\)
\(\Leftrightarrow5+2\left(\dfrac{5}{2}\right)^{2x}-7.\left(\dfrac{5}{2}\right)^x\le0\)
\(\Rightarrow1\le\left(\dfrac{5}{2}\right)^x\le\dfrac{5}{2}\)
\(\Rightarrow0\le x\le1\)
