a) Rút gọn biểu thức: \(P=\frac{{{x}^{2}}}{\left(...

B

- Hướng dẫn giải

Phương pháp giải:

a) Đặt điều kiện sau đó rút gọn biểu thức bằng các phép biến đổi: đặt nhân tử chung và sử dụng các hằng đẳng thức đáng nhớ.

b) Áp dụng kiến thức: Với \(a+b+c=0\) thì ta có: \({{\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)}^{2}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}.\)

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