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Câu hỏi: Cho 2 hàm số $y={{\log }_{2}}\left( x+2 \right) ({{C}_{1}})$ và $y={{\log }_{2}}x+1 \left( {{C}_{2}} \right)$. Goị $A,B$ lần lượt là giao điểm của $\left( {{C}_{1}} \right);\left( {{C}_{2}} \right)$ với trục hoành, $C$ là giao điểm của $\left( {{C}_{1}} \right)$ và $\left( {{C}_{2}} \right)$. Diện tích tam giác ABC bằng
A. $3 $ (đvdt)
B. $\dfrac{3}{4}$ (đvdt)
C. $\dfrac{3}{2}$ (đvdt)
D. $\dfrac{1}{2}$ (đvdt)
A. $3 $ (đvdt)
B. $\dfrac{3}{4}$ (đvdt)
C. $\dfrac{3}{2}$ (đvdt)
D. $\dfrac{1}{2}$ (đvdt)
* $\left( {{C}_{1}} \right)\cap \left( {{C}_{2}} \right)$
${{\log }_{2}}\left( x+2 \right)={{\log }_{2}}\left( x \right)+1\Leftrightarrow {{\log }_{2}}\left( x+2 \right)={{\log }_{2}}\left( 2x \right)$
$\Leftrightarrow x+2=2x\Leftrightarrow x=2\left( tm \right)$
$\Rightarrow \left( {{C}_{1}} \right)\cap \left( {{C}_{2}} \right)=C\left( 2;2 \right)$
* $\left( {{C}_{1}} \right)\cap Ox$
${{\log }_{2}}\left( x+2 \right)=0\Rightarrow A\left( -1;0 \right)$
* $\left( {{C}_{2}} \right)\cap Ox$
${{\log }_{2}}\left( x \right)+1=0\Rightarrow B\left( \dfrac{1}{2};0 \right)$
$\Rightarrow \overrightarrow{AB}\left( \dfrac{3}{2};0 \right);\overrightarrow{AC}\left( 3;2 \right)$
$\Rightarrow {{S}_{ABC}}=\dfrac{1}{2}\left| \overrightarrow{{{x}_{AB}}}.\overrightarrow{{{y}_{AC}}}-\overrightarrow{{{x}_{AC}}}.\overrightarrow{{{y}_{AB}}} \right|=\dfrac{3}{2}$ (đvdt).
${{\log }_{2}}\left( x+2 \right)={{\log }_{2}}\left( x \right)+1\Leftrightarrow {{\log }_{2}}\left( x+2 \right)={{\log }_{2}}\left( 2x \right)$
$\Leftrightarrow x+2=2x\Leftrightarrow x=2\left( tm \right)$
$\Rightarrow \left( {{C}_{1}} \right)\cap \left( {{C}_{2}} \right)=C\left( 2;2 \right)$
* $\left( {{C}_{1}} \right)\cap Ox$
${{\log }_{2}}\left( x+2 \right)=0\Rightarrow A\left( -1;0 \right)$
* $\left( {{C}_{2}} \right)\cap Ox$
${{\log }_{2}}\left( x \right)+1=0\Rightarrow B\left( \dfrac{1}{2};0 \right)$
$\Rightarrow \overrightarrow{AB}\left( \dfrac{3}{2};0 \right);\overrightarrow{AC}\left( 3;2 \right)$
$\Rightarrow {{S}_{ABC}}=\dfrac{1}{2}\left| \overrightarrow{{{x}_{AB}}}.\overrightarrow{{{y}_{AC}}}-\overrightarrow{{{x}_{AC}}}.\overrightarrow{{{y}_{AB}}} \right|=\dfrac{3}{2}$ (đvdt).
Đáp án C.