$1)\,\,x \ne 0,\,\,\,\,{9.4^{\frac{1}{x}}} + {5.6^{\frac{1}{x}}} = {4.9^{\frac{1}{x}}} \Leftrightarrow 9{\left( {\frac{2}{3}} \right)^{\frac{2}{x}}} + 5{\left( {\frac{2}{3}} \right)^{\frac{1}{x}}} = 4$Đặt $t = {\left( {\frac{2}{3}} \right)^{\frac{1}{x}}},\,\,t > 0\,$ ta có $9{t^2} + 5t - 4 = 0 \Leftrightarrow \left[ \begin{array}{l}t = - 1\\t = \frac{4}{9}\end{array} \right.$$t = - 1$ (loại) , $t = \frac{4}{9} = {\left( {\frac{2}{3}} \right)^2} = {\left( {\frac{2}{3}} \right)^{\frac{1}{x}}} \Leftrightarrow x = \frac{1}{2}$$2)$ ĐS: $x = - \frac{2}{3}$$3)$ ĐS: $x = - 1$$4)$ ĐS:$x = \frac{3}{2}$
$1)
ĐK:\,\,x \ne 0,\,\,\,\,
$Khi đó ta có${9.4^{\frac{1}{x}}} + {5.6^{\frac{1}{x}}} = {4.9^{\frac{1}{x}}}
$$\Leftrightarrow 9{\left( {\frac{2}{3}} \right)^{\frac{2}{x}}} + 5{\left( {\frac{2}{3}} \right)^{\frac{1}{x}}} = 4$
(chia cả 2 vế cho $
9^{\frac{1}{x}}$)Đặt $t = {\left( {\frac{2}{3}} \right)^{\frac{1}{x}}},\,\,t > 0\,$ ta có $9{t^2} + 5t - 4 = 0 \Leftrightarrow \left[ \begin{array}{l}t = - 1\\t = \frac{4}{9}\end{array} \right.$$t = - 1$ (loại) , $
\Rightarrow t = \frac{4}{9} = {\left( {\frac{2}{3}} \right)^2} = {\left( {\frac{2}{3}} \right)^{\frac{1}{x}}} \Leftrightarrow x = \frac{1}{2}$
(thỏa mãn điều kiện)Vậy PT có nghiệm $x=\frac{1}{2}$.$2)$
TXĐ: R.Phương trình đã cho tương đương với$2^{2x+4}.2-3^{x+\frac{7}{2}}.3=3^{x+\frac{7}{2}}-2^{2x+4}$$\Leftrightarrow 3.(2^{2x}.16)=3^x.3^\frac{7}{2}.(3+1)$$\Leftrightarrow \frac{4^x}{3^x}=\frac{9.\sqrt3.4}{3.16}=\left (\frac{3}{4}\right )^\frac{2}{3}\Rightarrow x=-\frac{2}{3}$ĐS: $x = - \frac{2}{3}$$3)$ Đ
K: $x\neq 0.$Khi đó phương trình đã cho tương đương$5.\left ( 5^\frac{1}{x} \right )^2+3.2^\frac{1}{x}.5^\frac{1}{x}-2.\left (2^\frac{1}{x} \right )^2$$\Leftrightarrow 5.\left[ {\left (\frac{5}{2}\right )^\frac{1}{x}} \right]^2+3.\left (\frac{5}{2} \right )^\frac{1}{x}-2=0(*)$.Đặt $t=
\left (\frac{5}{2}\right )^\frac{1}{x} >0$ thì (*) trở thành$5t^2+3t-2=0\Leftrightarrow \left[\begin{array}{l} t=-1(L)\\ t=\frac{2}{5}(TM) \end{array} \right.$$\Rightarrow t=\frac{2}{5}=
\left (\frac{5}{2}\right )^{-1} \Rightarrow x=-1.$ĐS: $x = - 1$$4)$
TXĐ: R.Phương trình đã cho tương đương với $5^{(x-\frac{1}{2})+1}-9^{(x-1)+1}=9^{x-1}-5^{x-\frac{1}{2}}$$\Leftrightarrow
5^{x-\frac{1}{2}}(5+1)=
9^{x-1} (9+1)$$\Leftrightarrow \frac{5^x.6}{\sqrt5}=\frac{9^x.10}{9}$$\Leftrightarrow \left (\frac{5}{9}\right )^x=\frac{5\sqrt5}{9.3}=\left (\frac{5}{9} \right )^\frac{3}{2}\Rightarrow x=\frac{3}{2}$.ĐS:$x = \frac{3}{2}$