Đáp án đúng: B
Giải chi tiết:a) Điều kiện: \(x\ne -y,\ \ x\ne -1,\ \ y\ne 1.\)
\(\begin{align} & \,P=\frac{{{x}^{2}}}{\left( x+y \right)\left( 1-y \right)}-\frac{{{y}^{2}}}{\left( x+y \right)\left( 1+x \right)}-\frac{{{x}^{2}}{{y}^{2}}}{\left( 1+x \right)\left( 1-y \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{x+y}\left( \frac{{{x}^{2}}}{1-y}-\frac{{{y}^{2}}}{1+x} \right)-\frac{{{x}^{2}}{{y}^{2}}}{\left( 1+x \right)\left( 1-y \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{x+y}.\frac{{{x}^{2}}+{{x}^{3}}-{{y}^{2}}+{{y}^{3}}}{\left( 1-y \right)\left( 1+x \right)}-\frac{{{x}^{2}}{{y}^{2}}}{\left( 1+x \right)\left( 1-y \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{x+y}.\frac{\left( x-y \right)\left( x+y \right)+\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)}{\left( 1-y \right)\left( 1+x \right)}-\frac{{{x}^{2}}{{y}^{2}}}{\left( 1+x \right)\left( 1-y \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{x+y}.\frac{\left( x+y \right)\left( x-y+{{x}^{2}}-xy+{{y}^{2}} \right)}{\left( 1-y \right)\left( 1+x \right)}-\frac{{{x}^{2}}{{y}^{2}}}{\left( 1+x \right)\left( 1-y \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\frac{x-y+{{x}^{2}}-xy+{{y}^{2}}}{\left( 1-y \right)\left( 1+x \right)}-\frac{{{x}^{2}}{{y}^{2}}}{\left( 1+x \right)\left( 1-y \right)} \\ & \,\,\,\,\,\,\,\,\,\,\,=\frac{x-y+{{x}^{2}}-xy+{{y}^{2}}-{{x}^{2}}{{y}^{2}}}{\left( 1-y \right)\left( 1+x \right)} \\ & \,\,\,\,\,\,\,\,\,\,=\frac{\left( x-y \right)+x\left( x-y \right)+{{y}^{2}}\left( 1-{{x}^{2}} \right)}{\left( 1-y \right)\left( 1+x \right)} \\ & \,\,\,\,\,\,\,\,\,\,=\frac{\left( x-y \right)\left( x+1 \right)+{{y}^{2}}\left( 1-x \right)\left( x+1 \right)}{\left( 1-y \right)\left( 1+x \right)}=\frac{\left( x+1 \right)\left( x-y+{{y}^{2}}-x{{y}^{2}} \right)}{\left( 1-y \right)\left( 1+x \right)} \\ & \,\,\,\,\,\,\,\,\,=\frac{x\left( 1-{{y}^{2}} \right)-y\left( 1-y \right)}{1-y}=\frac{\left( 1-y \right)\left( x+xy-y \right)}{1-y} \\ & \,\,\,\,\,\,\,\,\,=x+xy-y. \\ \end{align}\)
b) Chứng minh rằng: \(\sqrt{1+\frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}}+\sqrt{1+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}}+.....+\sqrt{1+\frac{1}{{{2017}^{2}}}+\frac{1}{{{2018}^{2}}}}<2018\)
Ta có: \({{\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)}^{2}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}+2.\left( \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac} \right)=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}+\frac{2\left( a+b+c \right)}{abc}.\)
\(\Rightarrow a+b+c=0\) ta có: \({{\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)}^{2}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}\)
Ta có: \(1+1+\left( -2 \right)=0\) nên \(\sqrt{1+\frac{1}{{{1}^{2}}}+\frac{1}{{{\left( -2 \right)}^{2}}}}=\sqrt{1+\frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}}=\sqrt{{{\left( 1+1-\frac{1}{2} \right)}^{2}}}=1+\frac{1}{2}.\)
Tương tự ta có:
\(\begin{align} & \sqrt{1+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}}=\sqrt{1+\frac{1}{{{2}^{2}}}+\frac{1}{{{\left( -3 \right)}^{2}}}}=1+\frac{1}{2}-\frac{1}{3} \\ & \sqrt{1+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}}=\sqrt{1+\frac{1}{{{3}^{2}}}+\frac{1}{{{\left( -4 \right)}^{2}}}}=1+\frac{1}{3}-\frac{1}{4} \\ & .................... \\ & .................... \\ & \sqrt{1+\frac{1}{{{2017}^{2}}}+\frac{1}{{{2018}^{2}}}}=\sqrt{1+\frac{1}{{{2017}^{2}}}+\frac{1}{{{\left( -2018 \right)}^{2}}}}=1+\frac{1}{2017}-\frac{1}{2018}. \\ & \Rightarrow VT=1+1+\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+.........+1+\frac{1}{2017}-\frac{1}{2018}=2018-\frac{1}{2018}<2018\,\,\left( dpcm \right). \\ \end{align}\)
Chọn B
