Giải phương trình sau:
a) $5^{2x+1}+7^{x+1}-175^x-35=0                 (1)$
b) $3.4^x-\frac{1}{3}.9^{x+2}=6.4^{x+1}-\frac{1}{2}.9^{x+1}  (2)$
c) $4^{x^2+x}+2^{1-x^2}=2^{(x+1)^2}+1                           (3)$
a) $(1) \Leftrightarrow 5^{2x+1}-35+7^{x+1}-25^x.7^x=0$
$\Leftrightarrow 5(25^x-7)-7^x(25^x-7)=0 \Leftrightarrow (5-7^x)(25^x-7)=0$
$\Leftrightarrow \left[ \begin{array}{l}
5-7^x=0\\
25^x-7=0\end{array} \right.\Leftrightarrow \left[ \begin{array}{l} 7^x=5 \\25^x=7 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x=\log_75 \\x=\log_{25}7 \end{array} \right.$

b) $(2)\Leftrightarrow 3.4^x-6.4^{x+1}=-\frac{1}{3}9^{x+2}-\frac{1}{2}9^{x+1}$
$\Leftrightarrow 4^x(3-24)=9^x(-27-\frac{9}{2})\Leftrightarrow 21.4^x=9^x.\frac{63}{2}\Leftrightarrow 4^x=\frac{3}{2}.9^x$
$\Leftrightarrow (\frac{4}{9})^x=\frac{3}{2}\Leftrightarrow (\frac{2}{3})^{2x}=(\frac{2}{3})^{-1}\Leftrightarrow x=-\frac{1}{2} $

c) $\Leftrightarrow 2^{2(x^2+x)}+2^{1-x^2}-2^{2(x^2+x)}.2^{1-x^2}-1=0$
$\Leftrightarrow 2^{2(x^2+x)}(1- 2^{1-x^2})-(1- 2^{1-x^2})=0\Leftrightarrow [2^{2(x^2+x)}-1](1- 2^{1-x^2})=0$
$\Leftrightarrow \left[ \begin{array}{l} 2^{2(x^2+x)}=1=2^0 \\2^{1-x^2}=1=2^0 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l} x^2+x=0 \\1-x^2=0 \end{array} \right.\Leftrightarrow \left[ \begin{array}{l}
x=0\\
x=-1\\
x=1\end{array} \right.
$

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