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b) Để \(\frac{6}{x+1}.\frac{x-1}{3}\)là một số nguyên =>\(\frac{6.\left(x-1\right)}{\left(x+1\right).3}\)phải là một số nguyên
Ta có:
\(\frac{6.\left(x-1\right)}{\left(x+1\right).3}=\frac{2\left(x-1\right)}{x+1}=\frac{2\left(x+1\right)-3}{x+1}\)=> Để \(\frac{6}{x+1}.\frac{x-1}{3}\)là một số nguyên thì 2(x+1)-3 phải chia hết cho x+1
=> 3 phải chia hết cho x+1
=> x+1 thuộc vào Ư(3)=(1;-1;3;-3)
Ta có bảng
| x+1 | 1 | -1 | 3 | -3 |
| x | 0 | -2 | 2 | -4 |
Vậy x=0;-2;2;-4 thì thỏa mãn yêu cầu đề bài
Ta có : P = \(\left|a-\frac{1}{2014}\right|+\left|a-\frac{1}{2016}\right|\)
Thay a = \(\frac{1}{2015}\)vào biểu thức P ,ta có :
\(\left|\frac{1}{2015}-\frac{1}{2014}\right|+\left|\frac{1}{2015}-\frac{1}{2016}\right|\)
\(=\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(=\frac{1}{2014}-\frac{1}{2016}\)
\(=\frac{2016-2014}{2014.2016}=\frac{2}{4060224}=\frac{1}{2030112}\)
Vậy P = \(\frac{1}{2030112}\)
\(\left(-3\right)^{2015}x\left(\frac{1}{3}\right)^{2015}+\left(0.25\right)^{2016}x4^{2016}\)
=\(\left(-3x\frac{1}{3}\right)^{2015}+\left(0.25x4\right)^{2016}\)
=\(\left(-1\right)^{2015}+1^{2016}\)
=\(-1+1\)
=\(0\)
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{101}{102}=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}=\frac{C}{D}\)
Ta có: \(D=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)(có 2015 số hạng)
\(D=\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)-2015\)
\(D=2016+\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}-2015\)
\(D=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+1=\frac{2016}{2}+\frac{2016}{3}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(D=2016\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=2016C\)
Vậy \(B=\frac{C}{D}=\frac{C}{2016C}=\frac{1}{2016}\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{102}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{101}{102}=\frac{1\cdot2\cdot3\cdot....\cdot101}{2\cdot3\cdot4\cdot....\cdot102}\)
\(A=\frac{1}{102}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\left(\frac{2015}{1}+1\right)+\left(\frac{2014}{2}+1\right)+...+\left(\frac{1}{2015}+1\right)+1}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{\frac{2016}{1}+\frac{2016}{2}+...+\frac{2016}{2015}+\frac{2016}{2016}}\)
\(B=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}}{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}=\frac{1}{2016}\)
a)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right).\\A = \left( {\frac{{30}}{{15}} + \frac{5}{{15}} - \frac{6}{{15}}} \right) - \left( {\frac{{105}}{{15}} - \frac{9}{{15}} - \frac{{20}}{{15}}} \right) - \left( {\frac{3}{{15}} + \frac{{25}}{{15}} - \frac{{60}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} - \left( {\frac{{ - 32}}{{15}}} \right)\\A = \frac{{29}}{{15}} - \frac{{76}}{{15}} + \frac{{32}}{{15}}\\A = \frac{{ - 15}}{{15}}\\A = - 1\end{array}\)
b)
\(\begin{array}{l}A = \left( {2 + \frac{1}{3} - \frac{2}{5}} \right) - \left( {7 - \frac{3}{5} - \frac{4}{3}} \right) - \left( {\frac{1}{5} + \frac{5}{3} - 4} \right)\\A = 2 + \frac{1}{3} - \frac{2}{5} - 7 + \frac{3}{5} + \frac{4}{3} - \frac{1}{5} - \frac{5}{3} + 4\\A = \left( {2 - 7 + 4} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right) + \left( { - \frac{2}{5} + \frac{3}{5} - \frac{1}{5}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
2A=\(\left(1+\frac{1}{3}\right)\)\(\left(1+\frac{1}{8}\right)\)\(\left(1+\frac{1}{15}\right)\)\(.......\)\(\left(1+\frac{1}{4064255}\right)\)
2A = \(\frac{4}{3}\)\(.\)\(\frac{9}{8}\)\(.\)\(\frac{16}{15}\)\(......\)\(\frac{4064256}{4064255}\)
2A = \(\frac{2.2}{1.3}\)\(.\)\(\frac{3.3}{2.4}\)\(.\)\(\frac{4.4}{3.5}\)\(......\)\(\frac{2016.2016}{2015.2017}\)
2A = \(\frac{2.3.4....2016}{1.2.3.....2015}\)\(.\)\(\frac{2.3.4....2016}{3.4.5....2017}\)
2A = \(\frac{2016}{1}\)\(.\)\(\frac{2}{2017}\)
2A = \(\frac{4032}{2017}\)
A = \(\frac{4032}{2017}\)\(:2\)
A = \(\frac{2016}{2017}\)
Thay \(a=\frac{1}{2015}\) vào biểu thức P ta được:
\(P=\left|\frac{1}{2015}-\frac{1}{2014}\right|+\left|\frac{1}{2015}-\frac{1}{2016}\right|\)
Ta có: \(P=\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(P=\frac{1}{2014}-\frac{1}{2016}\)
\(P=\frac{2016-2014}{2014.2016}=\frac{2}{2014.2016}\)
\(P=\frac{1}{1007.2016}=\frac{1}{2030112}\)