x là gì vậy bạn

đây là 1 phân thức

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

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

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

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 \(x-2004=0\)

\(\Rightarrow x=2004\)

Bài 2 : 1 + ( -2 ) + 3 + ( -4 )  + ... + 2015

= [ 1 + ( -2 ) ] + [ 3 + ( -4 ) ] + ... + 2015

= -1 + -1 + ... + 2015

Có số các cặp số bằng ( -1 ) là : 

2014 : 2 = 1007 ( cặp ) 

= -1007 + 2015

= 1008

Bài 1 sai đề

1/15

giup minh voi, di ma

Ta có; \(\frac13+\frac16+\frac{1}{10}+\cdots+\frac{1}{\frac{x\left(x+1\right)}{2}}=\frac{2001}{2002}\)

=>\(\frac26+\frac{2}{12}+\frac{2}{20}+\cdots+\frac{2}{x\left(x+1\right)}=\frac{2001}{2002}\)

=>\(2\left(\frac16+\frac{1}{12}+\frac{1}{20}+\cdots+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2002}\)

=>\(2\left(\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2002}\)

=>\(2\left(\frac12-\frac{1}{x+1}\right)=\frac{2001}{2002}\)

=>\(1-\frac{2}{x+1}=\frac{2001}{2002}\)

=>\(\frac{2}{x+1}=1-\frac{2001}{2002}=\frac{1}{2002}\)

=>x+1=4004

=>x=4003(nhận)

\(\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}=\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}\) \(\left(dkxd:x\ne-2\right)\)

\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)

Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm

\(\Leftrightarrow6x+2-x=0\)

\(\Leftrightarrow5x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)

Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)

\(\dfrac{\left(x^2+x+1\right).\left(3x+1\right)}{x+2}=\dfrac{\left(x^2+x+1\right).x}{2\left(x+2\right)}\)

hay \(\left(x^2+x+1\right).\dfrac{3x+1}{x+2}=\left(x^2+x+1\right).\dfrac{x}{2\left(x+2\right)}\)

\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)

\(\Rightarrow3x^2-3^2-3x^2+15x=1\)

\(\Rightarrow3x^2-9-3x^2+15x=1\)

\(\Rightarrow-9+15x=1\)

\(\Rightarrow15x=-8\)

\(\Rightarrow x=\frac{-8}{15}\)