\(\dfrac{x-17}{1997}+\dfrac{x-21}{1993}+\dfrac{x+2}{1008}=4\)

\(\Leftrightarrow\dfrac{x-17}{1997}-1+\dfrac{x-21}{1993}-1+\dfrac{x+2}{1008}-2=0\)\(\Leftrightarrow\dfrac{x-2014}{1997}+\dfrac{x-2014}{1993}+\dfrac{x-2014}{1008}=0\) \(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{1993}+\dfrac{1}{1997}+\dfrac{1}{1008}\right)=0\)vì \(\dfrac{1}{1993}+\dfrac{1}{1997}+\dfrac{1}{1008}\ne0\Rightarrow x-2014=0\Rightarrow x=2014\)

\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}=\frac{x+7}{1997}+\frac{x+9}{1995}+\frac{x+11}{1993}\)

\(\Leftrightarrow\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)

\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)

\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}-\frac{x+2004}{1997}-\frac{x+2004}{1995}-\frac{x+2004}{1993}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\right)=0\)

\(\Leftrightarrow x+2004=0\) ( do \(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\ne0\))

\(\Leftrightarrow x=-2004\)

\(\frac{x+1}{2003}\)\(+\)\(\frac{x+3}{2001}\)\(+\)\(\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+\frac{x+9}{1995}\)\(+\frac{x+11}{1993}\)

\(\Leftrightarrow\)\(\frac{x+1}{2003}\)\(+1+\)\(\frac{x+3}{2001}\)\(+1+\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+1+\frac{x+9}{1995}\)\(+1+\frac{x+11}{1993}\)

\(\Leftrightarrow\frac{x+2004}{2003}\)\(+\frac{x+2004}{2001}\)\(+\frac{x+2004}{1999}\)\(-\frac{x+2004}{1997}\)\(-\frac{x+2004}{1995}\)\(-\frac{x+2004}{1993}\)\(=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)

\(\Leftrightarrow x+2004=0\)(vì tích kia có kết quả khác 0)

\(\Leftrightarrow x=-2004\)

Vậy PT có tập nghiệm S = {-2004}

\(\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)

<=>\(\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)

<=>\(\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)

Mà \(\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)< 0\)

=> X+2004=0

=>X=-2004

 Vì 1x 2 x 3 x 4 có tận cùng là 4 , 5 x 6 x 7 x8 x 9 có tận cùng là 0  mà 4 + 0 =4 nên 1991 x 1992 x 1993 x 1994 + 1995 x 1996 x1997 x 1998 x1999 có tận cung là 4

chữ số tận cùng là 6 đó bạn

\(1991.1992.1993.1994.1995.1996.1997.1998.1999\)

\(=\left(.....1\right).\left(.....2\right).\left(......3\right).\left(.......4\right).\left(.........5\right).\left(.......6\right).\left(......7\right).\left(......8\right).\left(......9\right)\)

\(=\left(......2\right).\left(.....3\right)\left(....4\right)\left(.....5\right)\left(....6\right)\left(.....7\right)\left(....8\right)\left(......9\right)\)

\(=\left(....6\right)\left(.....4\right)\left(....5\right)\left(....6\right)\left(....7\right)\left(.....8\right)\left(.....9\right)\)

\(=\left(....24\right)\left(....5\right)\left(....6\right)\left(....7\right)\left(....8\right)\left(....9\right)\)

\(=\left(.....0\right)\left(....6\right)\left(....7\right)\left(.....8\right)\left(.....9\right)\)

\(=\left(....0\right)\left(....7\right)\left(....8\right)\left(....9\right)\)

\(=\left(.....0\right)\left(....8\right)\left(...9\right)\)

\(=\left(...0\right)\left(....9\right)\)

\(=\left(....0\right)\)

Vậy 1991 x 1992 x 1993 x 1994 x 1995 x 1996 x 1997 x 1998 x 1999 có chữ số tận cùng là 0

\(\frac{x+6}{1999}+\frac{x+8}{1997}=\frac{x+10}{1995}+\frac{x+12}{1993}\)

\(\Leftrightarrow\frac{x+6}{1999}+1+\frac{x+8}{1997}+1=\frac{x+10}{1995}+1+\frac{x+12}{1993}+1\)

\(\Leftrightarrow\frac{x+2005}{1999}+\frac{x+2005}{1997}=\frac{x+2005}{1995}+\frac{x+2005}{1993}\)

\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)

\(\Leftrightarrow x+2005=0\left(\frac{1}{1999}+\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\ne0\right)\)

<=> x=-2005

Vậy x=-2005

bạn chỉ cần cộng mỗi phân số với 1 là xong!

Vd: x+6/1999 +1 +x+8/1997 +1 = x+10/1995 +1 +x+12/1993 +1

(không quen sử dụng cái phần mềm này lắm nên mình không làm nốt được)

`[x-17]/1998+[x-21]/1994+[x+1]/1008=4`

`<=>[x-17]/1998-1+[x-21]/1994-1+[x+1]/1008-2=0`

`<=>[x-2015]/1998+[x-2015]/1994+[x-2015]/1008=0`

`<=>(x-2015)(1/1998+1/1994+1/1008)=0`

  `=>x-2015=0`

`<=>x=2015`

\(\dfrac{x-17}{1998}+\dfrac{x-21}{1994}+\dfrac{x+1}{1008}\text{=}4\)

\(\Leftrightarrow\dfrac{x-17}{1998}+\dfrac{x-21}{1994}+\dfrac{x+1}{1008}-4\text{=}0\)

\(\Leftrightarrow\left(\dfrac{x-17}{1998}-1\right)+\left(\dfrac{x-21}{1994}-1\right)+\left(\dfrac{x+1}{1008}-2\right)\text{=}0\)

\(\Leftrightarrow\left(\dfrac{x-2015}{1998}\right)+\left(\dfrac{x-2015}{1994}\right)+\dfrac{x-2015}{1008}\text{=}0\)

\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{1998}+\dfrac{1}{1994}+\dfrac{1}{1008}\right)\text{=}0\)

\(\Leftrightarrow\left(x-2015\right)\text{=}0\)

\(\Leftrightarrow x\text{=}2015\)

\(vay...\)

Lời giải:
ĐKXĐ: $x\geq 1$
Đặt $\sqrt{x+1}=a; \sqrt{x-1}=b$ (ĐK: $a,b\geq 0$)

PT đã cho trở thành:

$\frac{a^2+b^2}{2}+ab=a+b+4$
$\Leftrightarrow a^2+b^2+2ab=2(a+b)+8$

$\Leftrightarrow (a+b)^2-2(a+b)-8=0$

$\Leftrightarrow (a+b-4)(a+b+2)=0$

Với $a\geq 0; b\geq 0$ thì $a+b+2\geq 2>0$

$\Rightarrow a+b-4=0$

$\Leftrightarrow a+b=4$

$\Leftrightarrow \sqrt{x+1}+\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x+1}=4-\sqrt{x-1}$

$\Rightarrow x+1=15+x-8\sqrt{x-1}$ (bp 2 vế)

$\Leftrightarrow 14=8\sqrt{x-1}$

$\Leftrightarrow x-1=(\frac{7}{4})^2=\frac{49}{16}$

$\Leftrightarrow x=\frac{65}{16}$ (tm)