dễ thôi mà

Áp dụng tỉ lệ thức, ta có:

\(\Leftrightarrow\frac{108x-4970}{2915}=\frac{92x-4970}{2115}\Rightarrow\left(108x-4970\right)2115=2915\left(92x-4970\right)\)

=>x=100

Ông Thắng: làm kiểu đó chưa gọi là đúng hoàn toàn đâu

$\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}$x−4555 +x−4753 =x−5545 +x−5347 

=> x = 100 

\(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)

\(\Leftrightarrow\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

\(\Leftrightarrow\frac{x-45-55}{55}+\frac{x-47-53}{47}-\frac{x-55-45}{45}-\frac{x-53-47}{47}=0\)

\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{47}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

\(\Leftrightarrow x-100=0\)

\(\Leftrightarrow x=100\)

Vậy pt có tập nghiệm S = { 100 }

a) x+1/2004 + 1 + x+2/2003 +1 - x+3/2002 +1 - x+4/2001 +1

=> x+2005/2004 + x+2005/2003 - x+2005/2002 - x+2005/2001=0

=> (x + 2005) ( 1/2004+1/2003 - 1/2002 - 1/2001) =0

ta thấy 1/2004+1/2003-1/2002-1/2001 # 0

=> x+2005=0 => x=-2005

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

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

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

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

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

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

a, Mình nghĩ là đề sai .

b, Ta có : \(\frac{x-45}{55}+\frac{x-47}{45}=\frac{x-55}{45}+\frac{x-53}{47}\)

=> \(\frac{x-45}{55}-1+\frac{x-47}{45}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)

=> \(\frac{x-45}{55}-\frac{55}{55}+\frac{x-47}{53}-\frac{53}{53}=\frac{x-55}{45}-\frac{45}{45}+\frac{x-53}{47}-\frac{47}{47}\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

=> \(x-100=0\)

=> \(x=100\)

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

c, Ta có : \(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)

=> \(\frac{2-x}{2010}-1=\frac{1-x}{2011}+\frac{-x}{2012}\)

=> \(\frac{2-x}{2010}+1=\frac{1-x}{2011}+1+\frac{-x}{2012}+1\)

=> \(\frac{2-x}{2010}+\frac{2010}{2010}=\frac{1-x}{2011}+\frac{2011}{2011}+\frac{-x}{2012}+\frac{2012}{2012}\)

=> \(\frac{2012-x}{2010}=\frac{2012-x}{2011}+\frac{2012-x}{2012}\)

=> \(\frac{2012-x}{2010}-\frac{2012-x}{2011}-\frac{2012-x}{2012}=0\)

=> \(\left(2012-x\right)\left(\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)

=> \(2012-x=0\)

=> \(x=2012\)

Vậy phương trình trên có tập nghiệm là \(S=\left\{2012\right\}\)

\(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)

\(\Leftrightarrow\left(\dfrac{x-45}{55}-1\right)+\left(\dfrac{x-47}{53}-1\right)=\left(\dfrac{x-55}{45}-1\right)+\left(\dfrac{x-53}{47}-1\right)\)

\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)

\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)

Do \(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\) nên x - 100 = 0 <=> x = 100

mấy câu này dễ mà :V câu a+c lấy mỗi phân số trừ cho 1 ra tử chung rút ra thì tính b+d thì cộng một tử chung rồi lại tính tiếp thôi