Biến đổi tương đương thôi!

\(\frac{x^2-2x+2004}{x^2}\ge\frac{2003}{2004}\)

\(\Leftrightarrow2004x^2+2.2004.x+2004^2\ge2003x^2\)

\(\Leftrightarrow x^2+2.2004.x+2004^2\ge0\)

\(\Leftrightarrow\left(x+2004\right)^2\ge0\)(Luôn đúng)

ai ra đề cho 1 lạy

HELP ME!

`(2-x)/2002-1=(1-x)/2003-x/2004`

`<=>(2-x)/2002-1+(x-1)/2003+x/2004=0`(chuyển  vế)

`<=>(2-x)/2002+1+(x-1)/2003-1+x/2004-1=0`

`<=>(2004-x)/2002+(x-2004)/2003+(x-2004)/2004=0`

`<=>(x-2004)(1/2003+1/2004-1/2002)=0`

`<=>x=2004` do `1/2003+1/2004-1/2002 ne 0`

Vậy `x=2004`

\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)

\(\Leftrightarrow x=-2005\)

a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)

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

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)

b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)

\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)

a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền

b,  - 1 hoặc + 1 vs mỗi phân số nha

\(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+35=2^5\)

\(pt\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+3=0\)

\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1+\dfrac{x+3}{2002}+1=0\)

\(\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{2004}{2004}+\dfrac{x+2}{2003}+\dfrac{2003}{2003}+\dfrac{x+3}{2002}+\dfrac{2002}{2002}=0\)

\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}+\dfrac{x+2005}{2002}=0\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\right)=0\)

\(\Rightarrow x+2005=0\). Do \(\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\ne0\)

\(\Rightarrow x=-2005\)

\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}+3=0\\ \Rightarrow\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\right)\\ \Rightarrow x=-2010\)

Câu hỏi của Lan Anh - Toán lớp 8 | Học trực tuyến, bạn sửa số 9 thành số 0 ở VP ở dòng 2 nhé

ta có \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)

<=>\(\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)

<=>\(\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)

<=>\(\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)

<=>\(\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)

Vì\(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\Rightarrow2004-x=0\Rightarrow x=2004\)

Vậy nghiệm của phương trình là x=2004

\(\Leftrightarrow\dfrac{2-x}{2002}=\dfrac{1-x}{2003}+1-\dfrac{x}{2004}\)

\(\Leftrightarrow\left(\dfrac{2-x}{2002}+1\right)=\left(\dfrac{1-x}{2003}+1\right)+\left(1-\dfrac{x}{2004}\right)\)

=>2004-x=0

=>x=2004