\(\frac{43-x}{57}+\frac{46-x}{54}=\frac{49-x}{51}+\frac{52-x}{48}\)

\(\Leftrightarrow\left(\frac{43-x}{57}+1\right)+\left(\frac{46-x}{54}+1\right)=\left(\frac{49-x}{51}+1\right)+\left(\frac{52-x}{48}+1\right)\)

\(\Leftrightarrow\frac{43-x+57}{57}+\frac{46-x+54}{54}=\frac{49-x+51}{51}+\frac{52-x+48}{48}\)

\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}=\frac{100-x}{51}+\frac{100-x}{48}\)

\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}-\left(\frac{100-x}{51}+\frac{100-x}{48}\right)=0\)

\(\Leftrightarrow\left(100-x\right)\left[\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)\right]=0\) (*)

Vì\(\frac{1}{57}< \frac{1}{51},\frac{1}{54}< \frac{1}{48}\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)< \left(\frac{1}{51}+\frac{1}{48}\right)\)

\(\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)< 0\)

Phương trình (*) xảy ra khi: \(100-x=0\Leftrightarrow x=100\)

Vậy phương trình có nghiệm duy nhất là x = 100

ta có: (59-x)/41  +(57-x)/43  +(55-x)/45  +(53-x)/47  +(51-x)/49 =-5

       <=>[(59-x)/41  +1 ]  +[(57-x)/43  +1]  +[(55-x)/45  +1]  +[(53-x)/47  +1]   +[(51-x)/49  +1] =0

      <=>(59-x-41)/41      + (57-x-43)/43     +(55-x-45)/45     +(53-x-47)/47      +(51-x-49)/49    =0

      <=>(100-x)/41   +   (100-x)/43       +    (100-x)/45     +(100-x)/47      +   (100-x)/49   =0

     <=>(100-x).( 1/41   +  1/43  +  1/45   +   1/47   +  1/49  )  =0

mà   (1/41  +  1/43  +  1/45  +  1/47  +  1/49) khác 0  nên 100-x =0 <=>x=100

vậy nghiệm của pt là x=100

a) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+2\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

Vậy x = -10

b) \(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)

\(\Rightarrow\left(\frac{x+43}{57}+1\right)+\left(\frac{x+46}{54}+1\right)=\left(\frac{x+49}{51}+1\right)+\left(\frac{x+52}{48}+1\right)\)

\(\Rightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

\(\Rightarrow\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

Mà \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)

\(\Rightarrow x+100=0\)

\(\Rightarrow x=-100\)

Vậy x = -100

a.\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

=>\(\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

<=> \(\frac{x+1+9}{9}+\frac{x+2+8}{8}=\frac{x+3+7}{7}+\frac{x+4+6}{6}\)

<=>\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

<=> \(\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

<=> \(\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

<=> x+10=0

<=> x=-10

Vậy tập nghiệm của phương trình trên là S=\(\left\{-10\right\}\)

b. \(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)

=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)<=>\(\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)​

<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)

<=>(x+100)\(\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)\)=0

<=>x+100=0

<=>x= -100

Vậy tập nghiệm của phương trình trên là S=\(\left\{-100\right\}\)

Phương trình đầu bài tương đương với 
\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)\(\Leftrightarrow\frac{x+43+57}{57}+\frac{x+46+54}{54}=\frac{x+49+51}{51}+\frac{x+52+48}{48}\)\(\Leftrightarrow\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

\(\Leftrightarrow\orbr{\begin{cases}x+100=0\\\frac{1}{57}+\frac{1}{54}=\frac{1}{51}+\frac{1}{48}\left(sai\right)\end{cases}\Leftrightarrow x+100=0\Leftrightarrow x=-100}\)

Vậy phương trình có nghiệm duy nhất là x=-100

<=> \(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

<=> \(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

<=> \(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

vi \(\frac{1}{57}< \frac{1}{51};\frac{1}{54}< \frac{1}{48}\Rightarrow\frac{1}{57}-\frac{1}{51}+\frac{1}{54}-\frac{1}{48}< 0\)

=> x+100=0 => x= -100

vay pt co nghiem \(x=-100\)

Ta thấy:\(\frac{x+43}{57}\)\(+\)\(\frac{x+46}{54}\)\(+\)\(2\)\(=\)\(\frac{x+49}{51}\)\(+\)\(\frac{x+52}{48}\)\(+\)\(2\)

\(\Rightarrow\)\(\frac{x+43}{57}\)\(+\)\(\frac{57}{57}\)\(+\)\(\frac{x+46}{54}\)\(+\)\(\frac{54}{54}\)\(=\)\(\frac{x+49}{51}\)\(+\)\(\frac{51}{51}\)\(+\)\(\frac{x+48}{52}\)\(+\)\(\frac{52}{52}\)

\(\Leftrightarrow\)\(\frac{x+100}{57}\)\(+\)\(\frac{x+100}{54}\)\(=\)\(\frac{x+100}{51}\)\(+\)\(\frac{x+100}{52}\)

\(\Leftrightarrow\)\((\)\(x+100)\)\((\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\()\)\(=\)\((x+100)\)\((\frac{1}{52}\)\(+\)\(\frac{1}{51})\)

\(\Leftrightarrow\)\((x+100)\)\((\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\(-\)\(\frac{1}{52}\)\(-\)\(\frac{1}{51}\)\()\)\(=\)\(0\)\((1)\)

Ta thấy: \(\frac{1}{57}\)< \(\frac{1}{52}\)

          \(\frac{1}{54}\)<\(\frac{1}{51}\)

\(\Rightarrow\)\(\frac{1}{57}\)\(+\)\(\frac{1}{54}\)< \(\frac{1}{52}\)\(+\)\(\frac{1}{51}\)

\(\Rightarrow\)\(\frac{1}{57}\)\(+\)\(\frac{1}{54}\)\(-\)\(\frac{1}{52}\)\(-\)\(\frac{1}{51}\)< 0 \((2)\)

Từ \((1)\)và \(\left(2\right)\)\(\Rightarrow\)\(x+100\)\(=0\)

\(\Leftrightarrow x=-100\)

Vậy phương trình có tập nghiệm \(x=-100\)

\(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\)

\(\Rightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\frac{51-x}{49}+1\)\(=-5+5\)

\(\Rightarrow\frac{59-x+49}{41}+\frac{57-x+43}{43}+\frac{55-x+45}{45}+\frac{53-x+47}{47}\)\(+\frac{51-x+49}{49}=0\)

\(\Rightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)

\(\Rightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)

Vì \(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\ne0\)

\(\Rightarrow100-x=0\)

\(\Rightarrow x=100\)

\(=\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\)

\(\frac{51-x}{49}+1=-5+5\)

đoạn này có 5 là do mik mượn 5 con 1 khi đó nha

\(=\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\)

\(\frac{100-x}{49}=0\)

\(=\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)

mà \(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}< 0\)

nên 100-x=0 

còn lại bn từ lm

<=>\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)

<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)

<=>\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)

=>x+100=0

<=>x=-100

k nha bạn

\(\Leftrightarrow\frac{37x+1648}{1026}=\frac{11x+556}{272}\Rightarrow\left(37x+1648\right)272=1026\left(11x+556\right)\)

<=>(37x+1648)272=272(37x+1648)

=>272(37x+1648)=1026(11x+556)

=>10064x+448256=11286x+570456

<=>-1222x=122200

=>x=122200:-1222

=>x=-100 ( dễ hiểu chưa hả )

\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)

\(\Leftrightarrow\)\(\frac{x+43}{57}+1+\frac{x+46}{54}=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)

\(\Leftrightarrow\)\(\frac{x+43}{57}+\frac{57}{57}+\frac{x+46}{54}+\frac{54}{54}=\frac{x+49}{51}+\frac{51}{51}+\frac{x+52}{48}+\frac{48}{48}\)

\(\Leftrightarrow\)\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)

\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)

Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)nên \(x+100=0\Leftrightarrow x=-100\)

Vậy \(x=-100\)