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Câu hỏi: Cho hai số thực dương $a$ và $b$ thỏa mãn ${{4}^{ab}}{{.2}^{a+b}}=\dfrac{8\left( 1-ab \right)}{a+b}.$ Giá trị lớn nhất của biểu thức $P=ab+2a{{b}^{2}}$ bằng
A. 3.
B. 1.
C. $\dfrac{\sqrt{5}-1}{2}.$
D. $\dfrac{3}{17}.$
A. 3.
B. 1.
C. $\dfrac{\sqrt{5}-1}{2}.$
D. $\dfrac{3}{17}.$
Ta có ${{4}^{ab}}{{.2}^{a+b}}=\dfrac{8\left( 1-ab \right)}{a+b}\Leftrightarrow {{2}^{2ab+a+b}}=\dfrac{8\left( 1-ab \right)}{a+b}$
$\Rightarrow 2ab+a+b={{\log }_{2}}\dfrac{8\left( 1-ab \right)}{a+b}={{\log }_{2}}\left[ 2\left( 1-ab \right) \right]-{{\log }_{2}}\left( a+b \right)+2$
$\Rightarrow a+b+{{\log }_{2}}\left( a+b \right)=2\left( 1-ab \right)+{{\log }_{2}}\left[ 2\left( 1-ab \right) \right]$
$\Rightarrow a+b=2\left( 1-ab \right)\Rightarrow a=\dfrac{2-b}{2b+1}$
$\Rightarrow P=b.\dfrac{2-b}{2b+1}+2{{b}^{2}}.\dfrac{2-b}{2b+1}=\dfrac{2b-{{b}^{2}}+4{{b}^{2}}-2{{b}^{3}}}{2b+1}=-\dfrac{2{{b}^{3}}-3{{b}^{2}}-2b}{2b+1}$
$=-\dfrac{b\left( b-2 \right)\left( 2b+1 \right)}{2b+1}=-{{b}^{2}}+2b=-{{\left( b-1 \right)}^{2}}+1\le 1.$
$\Rightarrow 2ab+a+b={{\log }_{2}}\dfrac{8\left( 1-ab \right)}{a+b}={{\log }_{2}}\left[ 2\left( 1-ab \right) \right]-{{\log }_{2}}\left( a+b \right)+2$
$\Rightarrow a+b+{{\log }_{2}}\left( a+b \right)=2\left( 1-ab \right)+{{\log }_{2}}\left[ 2\left( 1-ab \right) \right]$
$\Rightarrow a+b=2\left( 1-ab \right)\Rightarrow a=\dfrac{2-b}{2b+1}$
$\Rightarrow P=b.\dfrac{2-b}{2b+1}+2{{b}^{2}}.\dfrac{2-b}{2b+1}=\dfrac{2b-{{b}^{2}}+4{{b}^{2}}-2{{b}^{3}}}{2b+1}=-\dfrac{2{{b}^{3}}-3{{b}^{2}}-2b}{2b+1}$
$=-\dfrac{b\left( b-2 \right)\left( 2b+1 \right)}{2b+1}=-{{b}^{2}}+2b=-{{\left( b-1 \right)}^{2}}+1\le 1.$
Đáp án B.