Giải giúp em bài 92 này ạ 92. Giải mỗi phương trình sau:,$a)~0,5^{2x^2+x-1}=\frac14;$ $b)~2^{x^2-6x-\frac52}=16\sqrt2;$,$c)~27^{x^2-4x+4}=9^{x^2-4};$ $d)~0,05^{x-3}=(2\sqrt5)^x;$,$e)~\log_33(x-2)=-1;$ $g)~\log_5(x^2+1)+\log_{0,2}(4-5x-x^2)=0.$

Question

Accepted Answer

a)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
0,5^{2x^{2} +x-1} =\frac{1}{4}\
\Longrightarrow 0,5^{2x^{2} +x-1} =0,5^{2}\
\Longrightarrow 2x^{2} +x-1=2\
\Longrightarrow 2x^{2} +x-3=0\
\Longrightarrow 2x^{2} +3x-2x-3=0\
\Longrightarrow x.( 2x+3) -( 2x+3) =0\
\Longrightarrow ( x-1) .( 2x+3) =0\
\Longrightarrow \left[ \begin{array}{l l}
x-1=0 & \
2x+3=0 &
\end{array} ight.\
\Longrightarrow \left[ \begin{array}{l l}
x=1 & \
x=-\frac{3}{2} &
\end{array} ight.
\end{array}$
Vậy phương trình có nghiệm $\displaystyle x=1$ hoặc $\displaystyle x=-\frac{3}{2}$
c)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
27^{x^{2} -4x+4} =9^{x^{2} -4}\
\Longrightarrow \left( 3^{3} ight)^{x^{2} -4x+4} =\left( 3^{2} ight)^{x^{2} -4}\
\Longrightarrow 3^{3x^{2} -12x+12} =3^{2x^{2} -8}\
\Longrightarrow 3x^{2} -12x+12=2x^{2} -8\
\Longrightarrow x^{2} -12x+20=0\
\Longrightarrow x^{2} -10x-2x+20=0\
\Longrightarrow x.( x-10) -2.( x-10) =0\
\Longrightarrow ( x-10) .( x-2) =0\
\Longrightarrow \left[ \begin{array}{l l}
x-10=0 & \
x-2=0 &
\end{array} ight.\
\Longrightarrow \left[ \begin{array}{l l}
x=10 & \
x=2 &
\end{array} ight.
\end{array}$
Vậy phương trình có nghiệm x=10 hoặc x=2
e)
ĐKXĐ: x>2
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\log_{3} 3.( x-2) =-1\
\Longrightarrow \log_{3} 3+\log_{3}( x-2) =-1\
\Longrightarrow 1+\log_{3}( x-2) =-1\
\Longrightarrow \log_{3}( x-2) =-2\
\Longrightarrow x-2=\frac{1}{9}\
\Longrightarrow x=\frac{19}{9}( tm)
\end{array}$
Vậy phương trình có nghiệm $\displaystyle x=\frac{19}{9}$