giaỉ phương trình (x+1).căn (5x^2+2x-3)=5x^2+4x-5

Question

giaỉ  phương trình (x+1).căn (5x^2+2x-3)=5x^2+4x-5

Accepted Answer

$\displaystyle \begin{array}{{>{\displaystyle}l}}
( x+1)\sqrt{5x^{2} +2x-3} =5x^{2} +4x-5\
( x+1)\sqrt{( x+1)( 5x-3)} =5x^{2} +2x-3+2x-2\ ( *) \
\end{array}$
Đặt $\displaystyle \ a=\sqrt{x+1} ,\ b=\sqrt{5x-3} \ ( a,b\ \geqslant 0)$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
a^{2} =x+1,\ \ \ \ b^{2} =5x-3\
b^{2} -a^{2} =4( x-1)
\end{array}$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
( *) \ \Longleftrightarrow \ a^{2} .ab=a^{2} b^{2} +2.\frac{b^{2} -a^{2}}{4}\
\Longleftrightarrow a^{3} b=a^{2} b^{2} +\frac{b^{2} -a^{2}}{2}\
\Longleftrightarrow \ 2a^{2} b( a-b) =( b-a)( b+a)\
\Longleftrightarrow ( a-b)\left( 2a^{2} b+b+a ight) =0\
\Longleftrightarrow \left[ \begin{array}{l l}
a-b=0 & \
2a^{2} b+b+a=0 &
\end{array} ight.
\end{array}$
+ TH1: $\displaystyle a-b=0\Longleftrightarrow \ a=b$
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\Rightarrow \sqrt{x+1} =\sqrt{5x-3}\
x+1=5x-3\
4x-4=0
\end{array}$
$\displaystyle x=1\ $(Thử lại thỏa mãn)

+ TH2: $\displaystyle 2a^{2} b+b+a=0$
Do $\displaystyle a,b\geqslant 0\Rightarrow 2a^{2} b+b+a\geqslant 0\ $
Dấu "=" xảy ra $\displaystyle \Longleftrightarrow a=b=0\ $
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\Longleftrightarrow \sqrt{x+1} =\sqrt{5x-3} =0\ \
\Rightarrow \begin{cases}
x=-1 & \
x=\frac{3}{5} &
\end{cases}( loại)
\end{array}$