Giải các bất phương trình :

$\begin{array}{l}
1)\,\sqrt {{{\log }_9}\left( {3{x^2} + 4x + 2} \right)}  + 1 > {\log _3}\left( {3{x^2} + 4x + 2} \right)\\
2)\,\sqrt {8 + {2.2^{\sqrt {3 - x} }} - {4^{\sqrt {3 - x} }} + } {2^{\sqrt {3 - x}  + 1}} > 5
\end{array}$$\begin{array}{l}
\end{array}$
$1)$    Điều kiện:$\left\{ \begin{array}{l}
{\log _9}\left( {3{x^2} + 4x + 2} \right) \ge 0\\
3{x^2} + 4x + 2 > 0
\end{array} \right.$$ \Leftrightarrow 3{x^2} + 4x + 1 \ge 0$.
BPT đã cho tương đương
$\sqrt{{\log _9}\left( {3{x^2} + 4x + 2} \right)}+1>2{\log _9}\left( {3{x^2} + 4x + 2} \right)$
Đặt $t = {\log _9}\left( {3{x^2} + 4x + 2} \right)\geq $   ta có bpt đã cho trở thành: $\sqrt t  + 1 > 2t$
$\Leftrightarrow \sqrt t  > 2t - 1 \Leftrightarrow \left[ \begin{array}{l}
0 \le t < \frac{1}{2}\\
\left\{ \begin{array}{l}
t \ge \frac{1}{2}\\
t > {\left( {2t - 1} \right)^2}
\end{array} \right.
\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l}
t \ge \frac{1}{2}\\
t > {\left( {2t - 1} \right)^2}
\end{array} \right.\,\,\,\,\, \Leftrightarrow \frac{1}{2} \le t < 1$
Suy ra $0\, \le t < 1\,\,  $
$\Leftrightarrow 1 \leq 3{x^2} + 4x + 2 < 9$
$\Leftrightarrow \left\{ \begin{array}{l} 3x^2+4x+1\geq 0\\ 3x^2+4x-7<0 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} (3x+1)(x+1)\geq 0\\ (x-1)(3x+7)<0 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x\geq -\frac{1}{3}\\ x\leq -1 \end{array} \right.\\ -\frac{7}{3}<x<1 \end{array} \right.\Rightarrow x\in \left( -{\frac{7}{3}};-1\right]\cup \left[ {-\frac{1}{3};1} \right)$
Vậy hệ đã cho có nghiệm $x\in \left( -{\frac{7}{3}};-1\right]\cup \left[ {-\frac{1}{3};1} \right)$
  
$2)$   
Điều kiện: $\left\{ \begin{array}{l} 3-x>0\\ 8+2.2^\sqrt{3-x}\geq 4^\sqrt{3-x} \end{array} \right.$
Đặt $t=2^\sqrt{3-x}\Rightarrow \left\{ \begin{array}{l} t>0\\ t^2-2t-8\leq 0 \end{array} \right.\Rightarrow 0<t\leq 4$
Khi đó BPT đã cho tương đương
$\sqrt{8+2t-t^2}+2t>5$
$\sqrt{8+2t-t^2}>5-2t$
$\Leftrightarrow \left[ \begin{array}{l} 5-2t<0\\ \left\{ \begin{array}{l} 5-2t\geq 0\\ 8+2t-t^2>(5-2t)^2 \end{array} \right. \end{array} \right.$
$\Leftrightarrow \left[\begin{array}{l} t> \frac{5}{2}\\ \left\{ \begin{array}{l} 0<t\leq \frac{5}{2}\\ 5t^2-22t+17<0 \end{array} \right.(I) \end{array} \right.$
Ta có $(I)\Leftrightarrow \left\{ \begin{array}{l} 0<t\leq \frac{5}{2}\\ (5t-17)(t-1)<0 \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} 0<t\leq \frac{5}{2}\\ 1<t<\frac{17}{5} \end{array} \right.\Rightarrow 1<t\leq \frac{5}{2}(TM)$
$\Leftrightarrow 0<\sqrt{3-x}\leq \log_2\frac{5}{2}$
$\Leftrightarrow 3-\log_2^2\frac{5}{2}\le x<3$
Vậy BPT đã cho có nghiệm: $3-\log_2^2\frac{5}{2}\le x<3$
dung roi ban oi..t>5/2 dau roi..minh lam quai ma k giong cua ban..hinu lam thieu roi –  giothienxung 06-05-13 08:50 AM
khoảng t>5/2 đâu rồi a td71220044 –  phankhoahoc002 10-11-12 11:42 PM
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