Google
Câu hỏi: Xét $x,y$ là hai só thực dương thỏa $1-\dfrac{1}{2}{{\log }_{2}}\left( x-y+2 \right)={{\log }_{2}}\left( \dfrac{x+1}{y}+1 \right).$ Tìm giá trị nhỏ nhất của biểu thức $P=\dfrac{x\left( y+1 \right)+10}{y}.$
A. 8.
B. 6.
C. 4.
D. 5.
A. 8.
B. 6.
C. 4.
D. 5.
Ta có $1={{\log }_{2}}\sqrt{x-y+2}+{{\log }_{2}}\dfrac{x+y+1}{y}={{\log }_{2}}\left( \dfrac{x+y+1}{y}.\sqrt{x-y+2} \right)$
$\Rightarrow \dfrac{x+y+1}{y}.\sqrt{x-y+2}=2\Rightarrow \left( x+y+1 \right)\left( \sqrt{x-y+2}-1 \right)=2y-\left( x+y+1 \right)$
$\Rightarrow \left( x+y+1 \right).\dfrac{x-y+1}{\sqrt{x-y+2}+1}+x-y+1=0\Rightarrow x-y+1=0\Rightarrow x=y-1$
$\Rightarrow P=\dfrac{\left( y-1 \right)\left( y+1 \right)+10}{y}=\dfrac{{{y}^{2}}+9}{y}=y+\dfrac{9}{y}\ge 2\sqrt{y.\dfrac{9}{y}}=6.$ Dấu "=" xảy ra
$\Leftrightarrow y=3;x=2.$
$\Rightarrow \dfrac{x+y+1}{y}.\sqrt{x-y+2}=2\Rightarrow \left( x+y+1 \right)\left( \sqrt{x-y+2}-1 \right)=2y-\left( x+y+1 \right)$
$\Rightarrow \left( x+y+1 \right).\dfrac{x-y+1}{\sqrt{x-y+2}+1}+x-y+1=0\Rightarrow x-y+1=0\Rightarrow x=y-1$
$\Rightarrow P=\dfrac{\left( y-1 \right)\left( y+1 \right)+10}{y}=\dfrac{{{y}^{2}}+9}{y}=y+\dfrac{9}{y}\ge 2\sqrt{y.\dfrac{9}{y}}=6.$ Dấu "=" xảy ra
$\Leftrightarrow y=3;x=2.$
Đáp án B.