Hỏi 1 câu đề thi thử thpt quốc gia

T

taythuyanh11

ĐKXĐ: $x \ge 0; 1 \le y \le 6$

Xét phương trình (1):

$x^2+xy-2y^2+3y-1=\sqrt{y-1}-\sqrt{x}$

$\iff (x+1-y)(x-1+2y)=\dfrac{y-1-x}{\sqrt{y-1}+\sqrt{x}}$

$\iff \begin{bmatrix}
y=x+1& \\
1-x-2y=\dfrac{1}{\sqrt{y-1}+\sqrt{x}} (vl)
\end{bmatrix}$

Thay $y=x+1$ vào pt (2):

$3\sqrt{5-x}+3\sqrt{5x-4}=2x+7$

$\iff 3\sqrt{5-x}-(7-x)+3\sqrt{5x-4}-3x=0$

$\iff \dfrac{-(x^2-5x+4)}{3\sqrt{5-x}+(7-x)}+\dfrac{-3(x^2-5x+4)}{\sqrt{5x-4}+x}=0$

$\iff \begin{bmatrix}
x^2-5x+4=0 & \\
\dfrac{1}{3\sqrt{5-x}+(7-x)}+\dfrac{3}{\sqrt{5x-4}+x}=0(vl) &
\end{bmatrix}$

$\iff \begin{bmatrix}
x=1 \rightarrow y=2 & \\
x=4 \rightarrow y=5&
\end{bmatrix}$
 
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