Áp dụng: `(y - x)/(x. y) = 1/x - 1/y`
Bài `1`:
`a_)`
`-1/4 : -3/4 + 1/2 < x < 7/8 - 1/2 : -5/6`
`=> 1/3 + 1/2 < x < 7/8 + 3/5`
`=> 5/6 < x < 59/40`
Vì `0 < 5/6, 59/40 < 80/40 = 2`
`=> 0 < x < 2` mà `x in mathbb Z`
`=> x = 1`
Vậy `x = 1`
`b_)`
`5/(1. 6) + 5/(6. 11) + ... + 5/((5x + 1)(5x + 6)) = 2010/2011`
`=> 1 - 1/6 + 1/6 - 1/11 + ... + 1/(5x + 1) - 1/(5x + 6) = 2010/2011`
`=> 1 - 1/(5x + 6) = 2010/2011`
`=> 1/(5x + 6) = 1 - 2010/2011`
`=> 1/(5x + 6) = 1/2011`
`=> 5x + 6 = 2011`
`=> 5x = 2005`
`=> x = 401`
Vậy `x = 401`
Bài `2`:
`A = 1/(2. 3) + 1/(3. 4) + ... + 1/(99. 100)`
`= 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100`
`= 1/2 - 1/100`
`= 49/100`
Vậy `A = 49/100`
`B = 5/(1. 4) + 5/(4. 7) + ... + 5/(100. 103)`
`= 5. (1/(1. 4) + 1/(4. 7) + ... + 1/(100. 103))`
`= 5. 1/3. (3/(1. 4) + 3/(4. 7) + ... + 3/(100. 103))`
`= 5/3. (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`
`= 5/3. (1 - 1/103)`
`= 5/3. 102/103`
`= 170/103`
Vậy `B = 170/103`
`C = 1/15 + 1/35 + ... + 1/2499`
`= 1/(3. 5) + 1/(5. 7) + ... + 1/(49. 51)`
`= 1/2. (2/(3. 5) + 2/(5. 7) + ... + 2/(49. 51))`
`= 1/2. (1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/51)`
`= 1/2. (1/3 - 1/51)`
`= 1/2. 16/51`
`= 8/51`
Vậy `C = 8/51`
Bài `3`:
`a_)`
`A = 3/(x - 1) (x ne 1)`
Để `A` nguyên thì `3/(x - 1)` nguyên
`=> 3 vdots x - 1`
`=> x - 1 in Ư(3) = {-1; 1; -3; 3}`
`=> x in {0; 2; -2; 4}`
Vậy `x in {0; 2; -2; 4}`
`b_)`
`B = (x - 2)/(x - 3) (x ne 3)`
Để `B` nguyên `(x - 2)/(x - 3)`
`=> x - 2 vdots x - 3`
`=> x - 3 + 1 vdots x - 3`
`=> 1 vdots x - 3`
`=> x - 3 in Ư(1) = {-1; 1}`
`=> x in {2; 4}`
Vậy `x in {2; 4}`
`c_)`
`C = (2x + 1)/(x - 3) (x ne 3)`
Để `C` nguyên thì `(2x + 1)/(x - 3)` nguyên
`=> 2x + 1 vdots x - 3`
`=> 2x - 6 + 7 vdots x - 3`
`=> 2(x - 3) + 7 vdots x - 3`
`=> 7 vdots x - 3`
`=> x - 3 in Ư(7) = {-1; 1; -7; 7}`
`=> x in {2; 4; -4; 10}`
Vậy `x in {2; 4; -4; 10}`
Bài `4`:
`a_)`
Đặt `ƯCLN(n + 1, 2n + 3) = d`
`=>` \(\left\{\begin{matrix}n + 1 \vdots d\\2n + 3 \vdots d\end{matrix}\right.\)
`=>` \(\left\{\begin{matrix}2(n + 1) = 2n + 2 \vdots d\\2n + 3 \vdots d\end{matrix}\right.\)
`=> 2n + 3 - (2n + 2) = 1 vdots d`
`=> d = +-1`
`=> ƯCLN(n + 1, 2n + 3) = +-1`
`=> (n + 1)/(2n + 3)` là phân số tối giản
`b_)`
Đặt `ƯCLN(2n + 3, 4n + 8) = d`
`=>` \(\left\{\begin{matrix}2n + 3 \vdots d\\4n + 8 \vdots d\end{matrix}\right.\)
`=>` \(\left\{\begin{matrix}2(2n + 3) = 4n + 6 \vdots d\\4n + 8 \vdots d\end{matrix}\right.\)
`=> 4n + 8 - (4n + 6) = 2 vdots d`
`=> d in {+-1; +-2}`
Vì `2n + 3` là số lẻ `=> d ne +-2`
`=> d = +-1`
`=> ƯCLN(2n + 3, 4n + 8) = +-1`
`=> (2n + 3)/(4n + 8)` là phân số tối giản
Bài `5`:
`a_)`
`A = (10^1992 + 1)/(10^1991 + 1)`
`= (10^1991. 10 + 10 - 9)/(10^1991 + 1)`
`= (10(10^1991 + 1) - 9)/(10^1991 + 1)`
`= (10(10^1991 + 1))/(10^1991 + 1) - 9/(10^1991 + 1)`
`= 10 - 9/(10^1991 + 1)`
`B = (10^1993 + 1)/(10^1992 + 1)`
`= (10^1992. 10 + 10 - 9)/(10^1992 + 1)`
`= (10(10^1992 + 1) - 9)/(10^1992 + 1)`
`= (10(10^1992 + 1))/(10^1992 + 1) - 9/(10^1992 + 1)`
`= 10 - 9/(10^1992 + 1)`
Vì `10^1991 + 1 < 10^1992 + 1`
`=> 9/(10^1991 + 1) > 9/(10^1992 + 1)`
`=> 10 - 9/(10^1991 + 1) < 10 - 9/(10^1992 + 1)`
`=> A < B`
Vậy `A < B`
`b_)`
`C = (2010^2008 + 1)/(2010^2009 + 1)`
`=> 2010C = (2010^2009 + 2010)/(2010^2009 + 1)`
`= (2010^2009 + 1)/(2010^2009 + 1) + 2009/(2010^2009 + 1)`
`= 1 + 2009/(2010^2009 + 1)`
`D = (2010^2007 + 1)/(2010^2008 + 1)`
`=> 2010D = (2010^2008 + 2010)/(2010^2008 + 1)`
`= (2010^2008 + 1)/(2010^2008 + 1) + 2009/(2010^2008 + 1)`
`= 1 + 2009/(2010^2008 + 1)`
Vì `2010^2009 + 1 > 2010^2008 + 1`
`=> 2009/(2010^2009 + 1) < 2009/(2010^2008 + 1)`
`=> 1 + 2009/(2010^2009 + 1) < 1 + 2009/(2010^2008 + 1)`
`=> 10C < 10D`
`=> C < D`
Vậy `C < D`
