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

\(\Rightarrow x^2+3x-1=3x^2\Leftrightarrow3x-1=2x^2\Leftrightarrow2x^2-3x+1=0\Leftrightarrow x^2-\frac{3}{2}x+\frac{1}{2}=0\)

đến đây là pt bậc 2

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

\(=\left(2.\left(-\frac{1}{2}\right)+1\right)^2-3\left(-\frac{1}{2}-1\right)^2-\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}-1\right)\)

\(=-3\left(-\frac{9}{4}\right)-\frac{1}{2}.\left(-\frac{3}{2}\right)\)

\(=\frac{27}{4}+\frac{3}{4}=\frac{31}{4}\)

còn đâu tự lm nha ! 

\(x\left(x-y\right)+y\left(y-x\right)=x\left(x-y\right)-y\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y\right)=\left(x-y\right)^2=\left(124-24\right)^2=100^2=10000\)

\(\left(x-3\right)^2-\left(x+1\right)^3+12x\left(x-1\right)=\frac{49}{4}-\frac{1}{8}+\frac{\left(-6\right).\left(-3\right)}{2}\)

\(=\frac{97}{8}+9=\frac{169}{8}\)

X(X-Y)+Y(Y-X)=X² -XY +Y² -XY=(X-Y)² =(124-24)² =100² =10000

(x-3)² -(x+1)³ +12x(x-1)=x² -6x+9-x³ -3x² -3x-1+12x² -12x=-x³ +10x² -9x+8

(3x+5)(2x-1)+(4x-1)(3x+2)

=6x²-3x+10x-5+12x²+8x-3x-2

=18x²+12x-7

#hoctot :)

\(\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)

\(=6x^2-3+10x-5+12x^2+8x-3x-2\)

\(=18x^2+12x-7\)

1) tam giác DEF có MN//EF

=> \(\frac{DM}{ME}=\frac{DN}{NF}=>\frac{2}{2}=\frac{3,5}{NF}=>NF=\frac{3,5.2}{2}=3,5cm\)

2)tam giasc DEF cos KI//EF

=>\(\frac{DK}{KE}=\frac{DI}{IF}=\frac{3}{1}=\frac{4,2}{IF}=IF=\frac{1.4,2}{3}=1,4cm\)