`(x-2003)/16 +(x-1997)/11 +(x-1992)/9 +(x-1991)/7=10`

`<=>((x-2003)/16-1)+((x-1997)/11-2)+((x-1992)/9-3)+((x-1991)/7-4)=0`

`<=>(x-2019)/16+ (x-2019)/11 +(x-2019)/9+(x-2019)/7 =0`

`<=> (x-2019)(1/16+1/11+1/9+1/7)=0`

<=> x-2019=0`

`<=> x=2019`

\(\dfrac{x-2003}{16}+\dfrac{x-1997}{11}+\dfrac{x-1992}{9}+\dfrac{x-1991}{7}=10\)

\(\Leftrightarrow\left(\dfrac{x-2003}{16}-1\right)+\left(\dfrac{x-1997}{11}-2\right)+\left(\dfrac{x-1992}{9}-3\right)+\left(\dfrac{x-1991}{7}-4\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-2003}{16}-\dfrac{16}{16}\right)+\left(\dfrac{x-1997}{11}-\dfrac{22}{11}\right)+\left(\dfrac{x-1992}{9}-\dfrac{27}{9}\right)+\left(\dfrac{x-1991}{7}-\dfrac{28}{7}\right)=0\)

\(\Leftrightarrow\dfrac{x-2003-16}{16}+\dfrac{x-1997-22}{11}+\dfrac{x-1992-27}{9}+\dfrac{x-1991-28}{7}=0\)

\(\Leftrightarrow\dfrac{x-2019}{16}+\dfrac{x-2019}{11}+\dfrac{x-2019}{9}+\dfrac{x-2019}{7}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\dfrac{1}{16}+\dfrac{1}{11}+\dfrac{1}{9}+\dfrac{1}{7}\right)=0\)

\(\Leftrightarrow x-2019=0\) Vì \(\dfrac{1}{16}+\dfrac{1}{11}+\dfrac{1}{9}+\dfrac{1}{7}\ne0\)

\(\Leftrightarrow x=2019\)

Vậy phương trình có 1 nghiệm \(x=2019\)

\(\dfrac{x+8}{12}+\dfrac{x+9}{11}+\dfrac{x+10}{10}+3=0\\ \Leftrightarrow\dfrac{x+8}{12}+1+\dfrac{x+9}{11}+1+\dfrac{x+10}{10}+1=0\\ \Leftrightarrow\dfrac{x+20}{12}+\dfrac{x+20}{11}+\dfrac{x+20}{10}=0\\ \Leftrightarrow\left(x+20\right)\left(\dfrac{1}{12}+\dfrac{1}{11}+\dfrac{1}{10}\right)=0\\ \Leftrightarrow x+20=0\Leftrightarrow x=-20\\ KL:...\)

`<=>((x+8)/12+1)+((x+9)/11+1)+((x+10)/10+1)=0`

`<=>(x+20)/12+(x+20)/11+(x+20)/10=0`

`<=>(x+20)(1/12+1/11+1/10)=0`

Vì `1/12+1/11+1/10 ≠ 0`

`=>x+20=0`

`=>x=0-20`

`=>x=-20`

a/\(\dfrac{8}{x-8}+1+\dfrac{11}{x-11}+1=\dfrac{9}{x-9}+1+\dfrac{10}{x-10}+1\)

=>\(\dfrac{8+x-8}{x-8}+\dfrac{11+x-11}{x-11}=\dfrac{9+x-9}{x-9}+\dfrac{10+x-10}{x-10}\)

=>\(\dfrac{x}{x-8}+\dfrac{x}{x-11}-\dfrac{x}{x-9}-\dfrac{x}{x-10}=0\)

=>x.\(\left(\dfrac{1}{x-8}+\dfrac{1}{x-11}+\dfrac{1}{x-9}+\dfrac{1}{x-10}\right)=0\)

=>x=0

b/\(\dfrac{x}{x-3}-1+\dfrac{x}{x-5}-1=\dfrac{x}{x-4}-1+\dfrac{x}{x-6}-1\)

=>\(\dfrac{x-x+3}{x-3}+\dfrac{x-x+5}{x-5}-\dfrac{x-x+4}{x-4}-\dfrac{x-6+6}{x-6}=0\)

=>\(\dfrac{3}{x-3}+\dfrac{5}{x-5}-\dfrac{4}{x-4}-\dfrac{6}{x-6}=0\)

Đến đây thì bạn giải giống câu a

\(\dfrac{x+4}{8}+\dfrac{x+3}{9}=\dfrac{x+2}{10}+\dfrac{x+1}{11}\)

\(\Leftrightarrow\left(\dfrac{x+4}{8}+1\right)+\left(\dfrac{x+3}{9}+1\right)=\left(\dfrac{x+2}{10}+1\right)+\left(\dfrac{x+1}{11}+1\right)\)

\(\Leftrightarrow\dfrac{x+12}{8}+\dfrac{x+12}{9}-\dfrac{x+12}{10}-\dfrac{x+12}{11}=0\)

\(\Leftrightarrow\left(x+12\right)\left(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}\right)=0\)

\(\Leftrightarrow x=-12\)( do \(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}\ne0\))

\(\dfrac{x+4}{8}+\dfrac{x+3}{9}=\dfrac{x+2}{10}+\dfrac{x+1}{11}\)

\(\dfrac{x+4}{8}+1+\dfrac{x+3}{9}+1=\dfrac{x+2}{10}+1+\dfrac{x+1}{11}+1\)

\(\dfrac{x+12}{8}+\dfrac{x+12}{9}=\dfrac{x+12}{10}+\dfrac{x+12}{11}\)

\(\dfrac{x+12}{8}+\dfrac{x+12}{9}-\dfrac{x+12}{10}-\dfrac{x+12}{11}=0\)

\(\Rightarrow\left(x+12\right).\left(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}\right)=0\)

Vì \(\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}\ne0\) nên \(x+12=0\)

\(\Rightarrow x=-12\)

\(\Rightarrow\left(\dfrac{x+4}{8}+1\right)+\left(\dfrac{x+3}{9}+1\right)=\left(\dfrac{x+2}{10}+1\right)+\left(\dfrac{x+1}{11}+1\right)\\ \Rightarrow\dfrac{x+12}{8}+\dfrac{x+12}{9}=\dfrac{x+12}{10}+\dfrac{x+12}{11}\\ \Rightarrow\left(x+12\right)\left(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}\right)=0\\ \Rightarrow x+12=0\left(\dfrac{1}{8}+\dfrac{1}{9}-\dfrac{1}{10}-\dfrac{1}{11}\ne0\right)\\ \Rightarrow x=-12\)

M~~1/4

M=1/4

bài 1:

\(\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}\)

<=>\(\left(\dfrac{x-10}{1994}-1\right)+\left(\dfrac{x-8}{1996}+-1\right)+\left(\dfrac{x-6}{1998}-1\right)=\left(\dfrac{x-2002}{2}-1\right)+\left(\dfrac{x-2000}{4}-1\right)+\left(\dfrac{x-1998}{6}-1\right)\)

<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}\)

<=>\(\dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}=0\)

<=>(x-2004)\(\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}\right)\)

vì 1/1994+1/1996+1/1998-1/2-1/4-1/6 khác 0

nên x-2004=0=>x=2004

vyaj.......

bài 2:

\(\dfrac{x-85}{15}+\dfrac{x-74}{13}+\dfrac{x-67}{11}+\dfrac{x-64}{9}=10\)

<=>\(\left(\dfrac{x-85}{15}-1\right)+\left(\dfrac{x-74}{13}-2\right)+\left(\dfrac{x-67}{11}-3\right)+\left(\dfrac{x-64}{9}-4\right)=0\)

<=>\(\dfrac{x-100}{15}+\dfrac{x-100}{13}+\dfrac{x-100}{11}+\dfrac{x-100}{9}=0\)

<=>\(\left(x-100\right)\left(\dfrac{1}{15}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{9}\right)=0\)

vì 1/15+1/13+1/11+1/9 khác 0

=>x-100=0<=>x=100

Câu a:

=> √(√x-3)²=2

=>|√x-3|=2

√x-3=2 hoặc √x-3=-2

=> x=25 hoặc x=1

Câu b:

=> (x+2)/17+1+(x+4)/15+1+(x+6)/13+1-(x+8)/11-1-(x+10)/9-1-(x+12)/7-1=0

=> (x+19)/17+(x+19)/15+(x+19)/13-(x+19)/11-(x+19)/9-(x+19)/7=0

=>(x+19)(1/17+1/15+1/13-1/11-1/9-1/7)=0

Vì 1/17+1/15+1/13-1/11-1/9-1/7 khác 0 nên x+19=0 =>x=-19

Bạn gắng đọc nhé vì dùng dt tl nên không viết dc web này tệ qua

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)