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Câu hỏi: Cho hàm số ${y = \dfrac{{{{\left( {2x - 1} \right)}^2}}}{{{{\left( {x + 2} \right)}^3}}}}$. Gọi ${{{y}_{C\tilde{N}}}}$ là giá trị cực đại của hàm số, ${{{y}_{CT}}}$ là giá trị cực tiểu của hàm số. Tích ${{{y}_{C\tilde{N}}}.{{y}_{CT}}}$ bằng
A. ${\dfrac{{32}}{{135}}.}$
B. ${\dfrac{{11}}{2}.}$
C. ${\dfrac{{11}}{4}.}$
D. ${0}$.
A. ${\dfrac{{32}}{{135}}.}$
B. ${\dfrac{{11}}{2}.}$
C. ${\dfrac{{11}}{4}.}$
D. ${0}$.
TXĐ của hàm số: $D=\mathbb{R}\backslash \left\{ -2 \right\}.$
Ta có: $y'=\dfrac{4\left( 2x-1 \right){{\left( x+2 \right)}^{3}}3{{\left( 2x-1 \right)}^{2}}{{\left( x+2 \right)}^{2}}}{{{\left( x+2 \right)}^{6}}}$
$=\dfrac{\left( 2x-1 \right){{\left( x+2 \right)}^{2}}\left( 4x+8-6x+3 \right)}{{{\left( x+2 \right)}^{6}}}=\dfrac{\left( 2x-1 \right){{\left( x+2 \right)}^{2}}\left( 2x+11 \right)}{{{\left( x+2 \right)}^{6}}}=\dfrac{\left( 2x-1 \right)\left( 2x+11 \right)}{{{\left( x+2 \right)}^{4}}}.$
Suy ra $y'=0\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{1}{2} \\
& x=\dfrac{11}{2} \\
\end{aligned} \right..$
Bảng biến thiên:
Suy ra ${{y}_{CN}}=\dfrac{32}{135},{{y}_{CT}}=0.$
Vậy ${{y}_{CN}}.{{y}_{CT}}=0.$
Ta có: $y'=\dfrac{4\left( 2x-1 \right){{\left( x+2 \right)}^{3}}3{{\left( 2x-1 \right)}^{2}}{{\left( x+2 \right)}^{2}}}{{{\left( x+2 \right)}^{6}}}$
$=\dfrac{\left( 2x-1 \right){{\left( x+2 \right)}^{2}}\left( 4x+8-6x+3 \right)}{{{\left( x+2 \right)}^{6}}}=\dfrac{\left( 2x-1 \right){{\left( x+2 \right)}^{2}}\left( 2x+11 \right)}{{{\left( x+2 \right)}^{6}}}=\dfrac{\left( 2x-1 \right)\left( 2x+11 \right)}{{{\left( x+2 \right)}^{4}}}.$
Suy ra $y'=0\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{1}{2} \\
& x=\dfrac{11}{2} \\
\end{aligned} \right..$
Bảng biến thiên:
Suy ra ${{y}_{CN}}=\dfrac{32}{135},{{y}_{CT}}=0.$
Vậy ${{y}_{CN}}.{{y}_{CT}}=0.$
Đáp án D.