Question

Bài 2: Giải pt, bpt sau:

1) \(3^{x+2} + 9^{x+1} = 4\)

2) \(2^{2x+6} + 2^{x+7} - 17 > 0\)

3) \(9^x - 2. 3 ^{x+1} + 5 < 0\)

4) \(8^x \leq 4.(4-2^x)\)

Answer 1

1: \(3^{x+2}+9^{x+1}=4\)

=>\(3^{x+2}+3^{2x+2}=4\)

=>\(3^x\cdot9+3^{2x}\cdot9=4\)

=>\(9\cdot\left(3^x+3^{2x}\right)=4\)

=>\(3^x+3^{2x}=\dfrac{4}{9}\)(1)

Đặt \(a=3^x\left(a>0\right)\)

Phương trình (1) sẽ trở thành \(a+a^2=\dfrac{4}{9}\)

=>\(a^2+a-\dfrac{4}{9}=0\)

=>\(\left[{}\begin{matrix}a=\dfrac{1}{3}\left(nhận\right)\\a=-\dfrac{4}{3}\left(loại\right)\end{matrix}\right.\)

=>\(3^x=\dfrac{1}{3}\)

=>x=-1

2: \(2^{2x+6}+2^{x+7}-17>0\)

=>\(2^{2x}\cdot64+2^x\cdot128-17>0\)

=>\(64\left(2^{2x}+2^x\cdot2\right)>17\)

=>\(2^{2x}+2^x\cdot2>\dfrac{17}{64}\)(2)

Đặt \(b=2^x\left(b>0\right)\)

(2) sẽ trở thành \(b^2+b\cdot2>\dfrac{17}{64}\)

=>\(b^2+2b-\dfrac{17}{64}>0\)

=>\(\left[{}\begin{matrix}b>\dfrac{1}{8}\left(nhận\right)\\b< -\dfrac{17}{8}\left(loại\right)\end{matrix}\right.\)

=>\(2^x>\dfrac{1}{8}\)

=>x>-3

3: \(9^x-2\cdot3^{x+1}+5< 0\)

=>\(\left(3^x\right)^2-6\cdot3^x+5< 0\)

=>\(\left(3^x-1\right)\left(3^x-5\right)< 0\)

=>\(1< 3^x< 5\)

=>\(0< x< log_35\)