giải hệ phương trình
[TEX]\left{\begin{(12x-7).\sqrt{3x-2} +y.(4y^2+1)= 0 (1)}\\{ \sqrt{x-1} + \sqrt{3-x} =(x-1).(y^2-2) (2)} [/TEX]
Giải :
ĐKXĐ :
[TEX]\left{\begin{1 \leq x \leq 3}\\{\left[\begin{y \geq \sqrt{2}}\\{y \leq - \sqrt{2}} } [/TEX]
(1) \Leftrightarrow $(12x-7).\sqrt{3x-2} +y.(4y^2+1)= 0$
$(4(3x-2)+1).\sqrt{3x-2}=(-y)(4.(-y)^2 +1 )$
xét hàm số: $f(t) = (4t^2+1) t = 4t^3+t$
f(t)' = 12t^2+1 >0
do đó : $f( \sqrt{3x-2})=f(-y)$
\Leftrightarrow $\sqrt{3x-2}=(-y)$
xét giá trị y \leq - $\sqrt{2}$
\Leftrightarrow $y^2 = 3x-2$ thay vào pt (2) ta có :
$\sqrt{x-1} + \sqrt{3-x} =(x-1).(3x-4)$
$\sqrt{x-1} + \sqrt{3-x} =3x^2-7x+4$
$\sqrt{x-1}-1 + \sqrt{3-x}-1 -(3x^2-7x+2) = 0$
$\frac{x-2}{\sqrt{x-1}+1}-\frac{x-2}{\sqrt{3-x}+1}-(x-2)(3x-1)=0$
$(x-2)(\frac{1}{\sqrt{x-1}+1}-\frac{1}{\sqrt{3-x}+1}-(3x-1) )=0$
[TEX]\left[\begin{x=2}\\{\frac{1}{\sqrt{x-1}+1}-\frac{1}{\sqrt{3-x}+1}-(3x-1) =0 (VN )} [/TEX]
x = 2 \Rightarrow y =-2