1) ${x^2-3x+2=x^2-x-2x+2=x(x-1)-2(x-1)=(x-1)(x-2)}$
2) ${x^2+4y^2-4xy-4=(x^2-4xy+4y^2)^2-4=(x-2y-4)(x-2y+4)}$
Bài 2:
${x^3-6x^2+12x-8+8x^3+12x^2+6x+1-9x^3-27x^2-27x-9=16}$
${-21x^2-9x-16=-16 <=>-3x(7x+3)=0 <=>x=0 và x=\frac{-3}{7}}$
$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 3:\ Cho\ ( a-b)^{2} +( b-c)^{2} +( c-a)^{2} =0\ \\ \Longrightarrow \ \begin{cases} a-b=0 & \\ b-c=0 & \Longrightarrow \begin{cases} a=b & \\ b=c & \\ c=a & \end{cases}\\ c-a=0 & \end{cases}\\ \rightarrow \ a=b=c( \ dpcm\ ) \ \\ b) \ \ a^{2} +b^{2} +1=ab+a+b\ \\ \rightarrow \ a^{2} +b^{2} +1-ab-a-b=0\\ \rightarrow \ 2a^{2} +2b^{2} +2-2ab-2a-2b=0\ \\ \rightarrow \ \left( a^{2} -2a+1\right) +\left( b^{2} -2b+1\right) +\left( a^{2} -2ab+b^{2}\right) =0\ \\ \rightarrow \ ( a-1)^{2} +( b-1)^{2} +( a-b)^{2} =0\ \\ \Longrightarrow \begin{cases} a-1=0 & \\ b-1=0 & \Longrightarrow \begin{cases} a=1 & \\ b=1 & \\ a=b & \end{cases}\\ a-b=0 & \end{cases}\\ Vậy\ a=b=1\ \rightarrow \ dpcm\ \end{array}$
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