\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)

\(\Rightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)

\(\Rightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)

\(\Rightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}\right)=0\)

Dễ thấy: \(\dfrac{1}{1000}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}>0\Leftrightarrow x-2007=0\Leftrightarrow x=2007\)

\(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)

<=>\(\dfrac{x-1016}{1001}-1+\dfrac{x-13}{1002}-2+\dfrac{x+992}{1003}-3=\dfrac{x+995}{1004}-3+\dfrac{x-7}{1005}-2\)

<=>\(\dfrac{x-2017}{1001}+\dfrac{x-2017}{1002}+\dfrac{x-2017}{1003}=\dfrac{x-2017}{1004}+\dfrac{x-2017}{1005}\)

<=>\(\left(x-2017\right)\left(\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}-\dfrac{1}{1004}-\dfrac{1}{1005}\right)=0\)

vì 1/1001+1/1002+1/1003-1/1004-1/1005 khác 0 nên x-2017=0<=>x=2017

vậy..........

ĐẶT A = 1001.1002.1003. ... .2016

TÍCH CÓ MỘT SỐ THỪA SỐ CHIA HÊT CHO 3 NHƯ 1002, 1005,...

MÀ SỐ NÀO NHÂN VỚI 3 THÌ CŨNG CHIA HÊT CHO 3 ( không tin bạn cứ thử) (1)

SUY RA TÍCH TRÊN CHĂC CHẮN CHIA HÊT CHO 3 (2)

ĐẶT B = 1.3.5. ...2015 

TÍCH TRÊN CŨNG CÓ THỪA SỐ 3 NÊN NÓ SẼ CHIA HẾT CHO 3 (3)

TỪ (2),(3) SUY RA A CHIA HẾT CHO B

tao cũng chịu nốt

\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)

\(\Leftrightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)

\(\Leftrightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)

\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}=0\right)\)

\(\Leftrightarrow x-2007=0\)

\(\Leftrightarrow x=2007\)

.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)

.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)

.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)

.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)

.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)

.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)

.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)

.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)

.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)

.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)

.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)

.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)

.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)

.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)

.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)

.Suy ra \(x-1000=0\Leftrightarrow x=1000\)

cảm ơn

mình làm luộn 2 câu còn lại nhé ^^

.c,\(|x|-\frac{15}{2}=\frac{15}{4}\)

.\(< =>|x|=\frac{15}{4}+\frac{15}{2}=\frac{15}{4}+\frac{30}{4}=\frac{45}{4}\)

.\(< =>\orbr{\begin{cases}x=\frac{45}{4}\\x=-\frac{45}{4}\end{cases}}\)

.d,\(|\frac{3}{4}-x|+1=\frac{3}{2}\)

.\(< =>|\frac{3}{4}-x|=\frac{3}{2}-1=\frac{3}{2}-\frac{2}{2}=\frac{1}{2}\)

.\(< =>\orbr{\begin{cases}\frac{3}{4}-x=\frac{1}{2}\\\frac{3}{4}-x=-\frac{1}{2}\end{cases}< =>\orbr{\begin{cases}x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\\x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\end{cases}}}\)