Ta có:

\(P=\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}\)

\(P=\dfrac{1}{2014}-\dfrac{1}{2016}=\dfrac{2016-2014}{2014.2016}\)

\(P=\dfrac{2}{2014.2016}=\dfrac{1}{1007.2016}=\dfrac{1}{2030112}\)

Từ \(\dfrac{a}{d}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\Rightarrow\left(\dfrac{a}{c}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\left(\dfrac{a-b}{c-d}\right)^{2014}\left(1\right)\)

Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\left(\dfrac{a}{c}\right)^{2014}=\left(\dfrac{b}{d}\right)^{2014}=\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{a^{2014}+b^{2014}}{c^{2014}+d^{2014}}=\left(\dfrac{a-b}{c-d}\right)^{2014}\)

\(A=\dfrac{1}{100}-\dfrac{1}{100.99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{100.99}+\dfrac{1}{99.98}+\dfrac{1}{98.97}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{100}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{97.98}+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=\dfrac{1}{100}-\left(1-\dfrac{1}{100}\right)\)

\(=-\dfrac{49}{50}\)

a, Ta có: \(A=\dfrac{4x-7}{x-2}=\dfrac{4\left(x-2\right)+1}{x-2}=4+\dfrac{1}{x-2}\)

Với \(x\in Z\) thì \(x-2\in Z\)

Để A nguyên thì \(\dfrac{1}{x-2}\) nguyên 

\(\Rightarrow x-2\in U\left(1\right)\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

\(B=\dfrac{3x^2-9x+2}{x-3}=\dfrac{3x\left(x-3\right)+2}{x-3}=3x+\dfrac{2}{x-3}\)

Với \(x\in Z\Rightarrow x-3\in Z\)

Để B nguyên thì \(\dfrac{2}{x-3}\) nguyên \(\Rightarrow x-3\in U\left(2\right)=\left\{\pm1;\pm2\right\}\)

Do đó \(x=5;x=1;x=4;x=2\)

Vậy để B nguyên thì \(x\in\left\{5;1;4;2\right\}\)

b, Từ câu a suy ra để A<B cùng nguyên thì \(x=1\)

\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)

\(=\dfrac{a\left(-x-y\right)+b\left(-x-y\right)-a\left(b-x\right)+y\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)

\(=\dfrac{-ax-ay-bx-by-ab+ax+by-xy}{abxy\left(xy+ay+ab+by\right)}\)

\(=\dfrac{-ay-bx-ab-xy}{abxy\left(xy+ay+ab+by\right)}\)

\(=\dfrac{-xy+ay+ab+by}{abxy\left(xy+ay+ab+by\right)}=\dfrac{-1}{abxy}\)

Với \(a=\dfrac{1}{3};b=-2;x=\dfrac{3}{2};y=1\)

\(\Rightarrow A=\dfrac{-1}{\dfrac{1}{3}.\left(-2\right).\dfrac{3}{2}.1}=-1\)

a, Điện trở tương đương đoạn mạch:

\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{12.6}{12+6}=4\Omega\)

b, Cương độ dòng điện qua mạch chính:

\(I=\dfrac{U}{R_{tđ}}=\dfrac{12}{4}=3\left(A\right)\)

Cường độ dòng điện qua điện trở R₁:

\(I_1=\dfrac{U}{R_1}=\dfrac{12}{12}=1\left(A\right)\)

Cường độ dòng điện qua điện trở R₂:

\(I_2=I-I_1=3-1=2\left(A\right)\)

c, Nhiệt lượng toả ra trên đoạn mạch trong thời gian 10 phút:

\(Q=I^2Rt=3^2.4.600=21600\left(J\right)\)

b, Điện trở của:

Đèn 1: \(R_1=\dfrac{U^2_1}{P_1}=\dfrac{220^2}{100}=484\left(\Omega\right)\)

Đèn 2: \(R_2=\dfrac{U^2_2}{P_2}=\dfrac{220^2}{80}=605\left(\Omega\right)\)

Đoạn mạch: \(R_{td}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{484.605}{484+605}\simeq269\left(\Omega\right)\)

\(P=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}\right)\)

\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)

\(=2.\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2012}\right)\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{2012}+\dfrac{1}{2013}=S\)

Do đó \(\left(S-P\right)^{2013}=0\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=kb,=kd\)

\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\) và \(\dfrac{5c+3d}{5c-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)

Vậy \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)