Bài làm

\(\frac{x+19}{27}-\frac{x+17}{29}=\frac{x+15}{31}-\frac{x+13}{33}\)

\(\Leftrightarrow\left(\frac{x+19}{27}+1\right)-\left(\frac{x+17}{29}+1\right)=\left(\frac{x+15}{31}+1\right)-\left(\frac{x+13}{33}+1\right)\)

\(\Leftrightarrow\frac{x+46}{27}-\frac{x+46}{29}=\frac{x+46}{31}-\frac{x+46}{33}\)

\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}=\left(x+46\right).\frac{1}{31}-\left(x+46\right).\frac{1}{33}\)

\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}-\left(x+46\right).\frac{1}{31}+\left(x+46\right).\frac{1}{33}=0\)

\(\Leftrightarrow\left(x+46\right)\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)=0\)

Mà \(\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)>0\forall x\)

\(\Leftrightarrow x+46=0\)

\(\Leftrightarrow x=-46\)

Vậy phương trình trên có tập nghiệm S = { -46 }

# Học tốt #

\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Rightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)

\(\Rightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Dễ  thấy\(\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)>0\Rightarrow x-50=0\Rightarrow x=50\)

Vậy x = 50

Ta có 

\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Mà : \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\)

\(\Rightarrow x-50=0\)

\(\Rightarrow x=50\)

Vậy : \(x=50\)

\(\Leftrightarrow\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}-4=0\)

\(\Rightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

\(\Rightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Mà \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\)nên 

\(\Rightarrow x-50=0\Rightarrow x=50\)

Vậy x=50

             \(\frac{x-20}{7}+\frac{x-46}{33}+\frac{x-5}{8}+\frac{x+6}{19}=0\)

\(\Leftrightarrow\)\(\left(\frac{x-20}{7}+1\right)+\left(\frac{x-46}{33}+1\right)+\left(\frac{x-5}{8}-1\right)+\left(\frac{x+6}{19}-1\right)=0\)

\(\Leftrightarrow\)\(\frac{x-13}{7}+\frac{x-13}{33}+\frac{x-13}{8}+\frac{x-13}{19}=0\)

\(\Leftrightarrow\)\(\left(x-13\right)\left(\frac{1}{7}+\frac{1}{33}+\frac{1}{8}+\frac{1}{19}\right)=0\)

\(\Leftrightarrow\)\(x-13=0\)       (vi  1/7 + 1/33 + 1/8 + 1/19 \(\ne0\))

\(\Leftrightarrow\)\(x=13\)

Vậy phương trình có nghiệm là     \(x=13\)

pt <=> ( x-20/7 +1 ) + ( x-46/33 + 1 ) + ( x-5/8 - 1 ) + ( x+6/19 - 1 ) = 0

<=> x-13/7 + x-13/33 + x-13/8 + x-13/19 = 0

<=> (x-13).( 1/7+1/33+1/8+1/19 ) = 0

<=> x-13 = 0 ( vì 1/7+1/33+1/8+1/19 > 0 )

<=> x=13

Vậy x=13

Tk mk nha

\(\frac{x-20}{7}+\frac{x-46}{33}+\frac{x-5}{8}+\frac{x+6}{19}=0\)

\(\Leftrightarrow\left(\frac{x-20}{7}+1\right)+\left(\frac{x-46}{33}+1\right)+\left(\frac{x-5}{8}-1\right)+\left(\frac{x+6}{19}-1\right)=0\)

\(\Leftrightarrow\frac{x-13}{7}+\frac{x-13}{33}+\frac{x-13}{8}+\frac{x-13}{19}=0\)

\(\Leftrightarrow\left(x-13\right)\left(\frac{1}{7}+\frac{1}{33}+\frac{1}{8}+\frac{1}{19}\right)=0\)

Mà \(\frac{1}{7}+\frac{1}{33}+\frac{1}{8}+\frac{1}{19}\ne0\)

\(\Rightarrow x-13=0\Leftrightarrow x=13\)

\(pt\Leftrightarrow\frac{29-x}{21}+1+\frac{27-x}{23}+1+...=0\)

\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)

Do \(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}>0\) nên 50 - x = 0 hay x = 50.

pt<=>29-x/21+1+27-x/23+1+...=0

<=>50-x/21+50-x/23+50-x/25+50-x/27+50-x/29=0

<=>(50-x).(1/21+1/23+1/25+1/27+1/29)=0

Do 1/21+1/23+1/25+1/27+1/29>0 nên 50-x=0 hay x=50

b, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1994}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)

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

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)

=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)

=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)=0\)

=> \(x-2004=0\)

=> \(x=2004\)

Vậy phương trình có tập nghiệm là \(S=\left\{2004\right\}\)

a) Sửa đề: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)

Ta có: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)

\(\Leftrightarrow\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)

\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)

\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)

\(\Leftrightarrow\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)

Vì \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\)

nên x+36=0

hay x=-36

Vậy: x=-36

Ta cong them 1 vao moi phan thuc:

         \(\frac{X+29}{1971}+1+\frac{X+27}{1973}+1+\frac{X+25}{1975}+1=\frac{X+1971}{29}+1+\frac{X+1973}{27}+1+\frac{X+1975}{25}+1\) 

  \(\Leftrightarrow\frac{X+2000}{1971}+\frac{X+2000}{1973}+\frac{X+2000}{1975}=\frac{X+2000}{29}+\frac{X+2000}{27}+\frac{X+2000}{25}\) 

   \(\Leftrightarrow\frac{X+2000}{1971}+\frac{X+2000}{1973}+\frac{X+2000}{1975}-\frac{X+2000}{29}-\frac{X+2000}{27}-\frac{X+2000}{25}=0\)

    \(\Leftrightarrow\left(X+2000\right)\left(\frac{1}{1971}+\frac{1}{1973}+\frac{1}{1975}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}\right)=0\)

     Vi  \(\frac{1}{1971}+\frac{1}{1973}+\frac{1}{1975}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}\ne0\) nen    \(X+2000=0\Leftrightarrow X=-2000\)

Vay \(X=-2000\)

mk cũng đồng ý vs cách làm của Nguyễn Đức Tiến