\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)

\(\Rightarrow\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)\(\Rightarrow\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)

\(\Rightarrow\left(x-2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=\left(x-2004\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)

Với \(x-2004\ne0\)

\(\Rightarrow\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\left(KTM\right)\)

Với \(x-2004=0\)

\(\Rightarrow x=2004\)

bạn lên học 24/7 hỏi nha

\(\frac{2}{\left(x+3\right)\left(x+1\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}=\frac{2}{9}\)

\(\Rightarrow\frac{2}{x+1}-\frac{2}{x+3}+\frac{2}{x+3}-\frac{2}{x+5}+\frac{2}{x+5}-\frac{2}{x+7}=\frac{2}{9}\)

\(\frac{2}{x+1}-\frac{2}{x+7}=\frac{2}{9}\\ \Rightarrow\frac{2x+14-2x-2}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}\\ \Rightarrow\frac{12}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}=\frac{12}{54}\)

\(\Rightarrow\left(x+1\right)\left(x+7\right)=54\\ \Rightarrow x^2+8x-54=0\Rightarrow x=-4\pm\sqrt{70}\)

chết oy mk nhầm nha

\(\Rightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}=\frac{2}{9}\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+7}=\frac{2}{9}\)

\(\Rightarrow\frac{6}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}=\frac{6}{27}\\ \Rightarrow\left(x+1\right)\left(x+7\right)=27\)

\(\Rightarrow x^2+8x-20=0\Rightarrow\left(x-2\right)\left(x+10\right)=0\Rightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)

sorry bạn nha mk nhầm xíu

x-5/1990+x-15/1980+x-25/1970=x-1990/5+x-1980/15+x-1970/25

<=> (x-5/1990-1)+(x-15/1980-1)+(x-25/1970-1)=(x-1990/5-1)+(x-1980/15-1)+(x-1970/25-1)

<=> x-1995/1990+x-1995/1980+x-1995/1970=x-1995/5+x-1995/15+x-1995/25

<=> (x-1995)(1/1990+1/1980+1/1970-1/5-1/15-1/25)=0

<=> x-1995=0 

<=> x=1995

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

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

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

 \(\Leftrightarrow x-100=0\Leftrightarrow x=100\)

2. \(\Leftrightarrow\frac{x-5}{1990}+1+\frac{x-15}{1980}+1+\frac{x-25}{1970}=\frac{x-1990}{5}+1+\frac{x-1980}{15}+1+\frac{x-1970}{25}+1\)

   \(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)

   \(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)

  \(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)

  \(\Leftrightarrow x-1995=0\Leftrightarrow x=1995\)

câu 3 hình như sai đề

Sai đề không bạn?

a, Điều kiện: 3x - 2 ≥ 0 => 3x ≥ 2 => x ≥ 2/3

Ta có: |2x + 1| = 3x - 2

\(\Rightarrow\orbr{\begin{cases}2x+1=3x-2\\2x+1=2-3x\end{cases}}\Rightarrow\orbr{\begin{cases}2x-3x=-2-1\\2x+3x=2-1\end{cases}}\Rightarrow\orbr{\begin{cases}-x=-3\\5x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}(lọai)\end{cases}}\)

Vậy x = 3

b, \(\frac{5}{x}=\frac{x}{25}\)\(\Rightarrow x^2=5.25\)\(\Rightarrow x^2=125\)\(\Rightarrow\orbr{\begin{cases}x=5\sqrt{5}\\x=-5\sqrt{5}\end{cases}}\)

  a,|2x+1| = 3x-2      (1)

Ta có \(\left|2x+1\right|\ge0\forall x\)

=> 3x - 2 \(\ge0\)

\(\Rightarrow3x\ge2\)

\(\Rightarrow x\ge\frac{2}{3}>0\)

\(\Rightarrow2x>0\)

\(\Rightarrow2x+1>1>0\)

\(\Rightarrow\left|2x+1\right|=2x+1\)  (2)

Từ (1) và (2) => \(2x+1=3x-2\)

\(\Rightarrow3x-2x=1+2\)

\(\Rightarrow x=3\)

Vậy x = 3

b, \(\frac{5}{x}=\frac{x}{25}\)

\(\Rightarrow x^2=25.5=125\)

\(\Rightarrow\orbr{\begin{cases}x=\sqrt{25}\\x=-\sqrt{25}\end{cases}}\)

Vậy \(x\in\left\{\sqrt{25};-\sqrt{25}\right\}\)

P/ s: Câu a là làm theo cách ngu học của mình

Có sai thì thông cảm

minh cam on moi ng nheu !!!!!!!!