\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}=3\)

<=> \(\left(\frac{x-152}{2011}-1\right)+\left(\frac{x-2135}{28}-1\right)+\left(\frac{x-2039}{124}-1\right)=0\)

<=> \(\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)

<=> \(\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)

<=> \(x-2163=0\)(vì \(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\ne0\)

<=> \(x=2163\)

\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}=3\)

<=> \(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}-3=0\)

<=> \(\left(\frac{x-28-124}{2011}-1\right)+\left(\frac{x-124-2011}{28}-1\right)+\left(\frac{x-2011-28}{124}-1\right)=0\)

<=> \(\frac{x-28-124-2011}{2011}+\frac{x-124-2011-28}{28}+\frac{x-2011-28-124}{124}=0\)

<=> \(\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)

<=> \(\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)

Vì \(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\ne0\)

=> \(x-2163=0\)

=> \(x=2163\)

ta co (x-2163)(1/2011+1/28+1/124)=0

vi 1/2011+1/28+1/124khac0

x-2163=0

x=2163

Áp dụng tính chất của dãy tỉ số bằng nhau

Áp dụng tính chất của dãy tỉ số bằng nhau nha ^^^

a )

\(\Rightarrow\frac{x-100}{24}-1+\frac{x-98}{26}-1+\frac{x-96}{26}-1=0\)

\(\frac{x-124}{24}+\frac{x-124}{26}+\frac{x-124}{28}=0\)

\(\left(x-124\right)\left(\frac{1}{26}+\frac{1}{24}+\frac{1}{28}\right)=0\)

\(\Rightarrow x-124=0\Rightarrow x=124\)

Dung mà có làm bài lớp 5 à ta :D

\(\frac{x-28-124}{2011}+\frac{x-124-2011}{28}+\frac{x-2011-28}{124}\)\(=3\)

\(\Leftrightarrow\frac{x-152}{2011}+\frac{x-2135}{28}+\frac{x-2039}{124}=3\)\(\Leftrightarrow\left(\frac{x-152}{2011}-1\right)+\left(\frac{x-2135}{28}-1\right)+\left(\frac{x-2039}{124}-1\right)=3-1-1-1\)

\(\Leftrightarrow\frac{x-2163}{2011}+\frac{x-2163}{28}+\frac{x-2163}{124}=0\)

\(\Leftrightarrow\left(x-2163\right)\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)=0\)

        Vì \(\left(\frac{1}{2011}+\frac{1}{28}+\frac{1}{124}\right)>0\)

  \(\Rightarrow x-2163=0\)

\(\Leftrightarrow x=2163\)

          Vậy \(x=2163\)

b) (4x -1)² = (1-4x)⁴  (1)

Vì (1 - 4x) = (4x - 1)

\(\Rightarrow\)(1 - 4x)⁴ = [ -( 4x -1)⁴ ]

Vì (1-4x)⁴ = ( 4x - 1)⁴

Do đó (1) có dạng :

(4x - 1)²  = (4x - 1)²

Đặt 4x - 1 = x, ta có : 

x²  = x⁴

x²  ( 1 - x² ) = 0

\(\Rightarrow\)\(\orbr{\begin{cases}x=0\\x^2=\orbr{\begin{cases}1^2\\\left(-1\right)^2\end{cases}}\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=\orbr{\begin{cases}x=1\\x=-1\end{cases}}\end{cases}}\)

Thay x = 4x -1 = 0 

         x = \(\frac{1}{4}\)

• x = 1 \(\Leftrightarrow\) 4x - 1 = 1

                                 x = \(\frac{1}{2}\)

• x = -1 \(\Leftrightarrow\) 4x -1 = -1

                                   x = 0

Vậy  x = \(\frac{1}{2}\) hoặc x = 0

x = \(\frac{1}{4}\) nữa bạn nhé