Ta có:

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

\(\Rightarrow x^2=3x^2-5x+2\)

\(\Rightarrow3x^2-5x+2-x^2=0\)

\(\Rightarrow2x^2-4x-x+2=0\)

\(\Rightarrow\left(2x^2-4x\right)-\left(x-2\right)=0\)

\(\Rightarrow2x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)

Vậy x = 2 hoặc x = 1/2

Bạn tự trả lời luôn rồi hả 

(1/4-1).(1/5-1).....(1/2000-1).(1/2001-1)

=(-3/4).(-4/5).(-5/6).....(-1999/2000).(-2000/2001)

=-3.-4.-5....-1999.-2000/4.5.6...2000.2001

=-3/2001

\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

=> \(\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)

 \(\Leftrightarrow a^2cd+b^2cd-abc^2-abd^2=0\)

\(\Leftrightarrow a^2cd-abc^2+b^2cd-abd^2=0\)

\(\Leftrightarrow ac\left(ad-bc\right)+bd\left(bc-ad\right)=0\)

\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)

\(\Leftrightarrow\left(ad-bc\right)\left(ac-bd\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}ad-bc=0\\ac-bd=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}ad=bc\\ac=bd\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)   (DPCM)

trùng nhau