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Bài 6:

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)

⇔\(\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)⇔\(\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)

⇔\(\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)

⇔\(\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)

Vì \(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\)

\(\Rightarrow x-2014=0\)

⇔\(x=2014\).

-Vậy \(S=\left\{2014\right\}\).

Bài 7:

\(x^2-4x+y^2-6y+15=2\)

⇔\(x^2-4x+y^2-6y+15-2=0\)

⇔\(x^2-4x+y^2-6y+13=0\)

⇔\(\left(x^2-4x+4\right)+\left(y^2-6y+9\right)=0\)

⇔\(\left(x-2\right)^2+\left(y-3\right)^2=0\)

Vì \(\left(x-2\right)^2\ge0,\left(y-3\right)^2\ge0\).

\(\Rightarrow\)\(\left(x-2\right)^2=0\) và \(\left(y-3\right)^2=0\)

⇔\(x=2\) và \(y=3\).

Bài 8:

\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)

⇔\(\left(x^2+1\right)^2+2x\left(x^2+1\right)+x^2+x\left(x^2+1\right)+x^2=0\)

⇔\(\left(x^2+1+x\right)^2+x\left(x^2+1+x\right)=0\)

⇔\(\left(x^2+1+x\right)\left(x^2+1+x+x\right)=0\)

⇔\(x^2+x+1=0\) (vô nghiệm, bạn tự c/m) hay \(x^2+2x+1=0\)

⇔\(x^2+2x+1=0\)

⇔\(\left(x+1\right)^2=0\)

⇔\(x=-1\).

-Vậy \(S=\left\{-1\right\}\)

Câu 6: D

Câu 7: D

Câu 8: A

6C
7D
8D

8:

a: Xét tứ giác ADCF có

E là trungd diểm chung của AC và DF

=>ADCF là hbh

=>AD//CF và AD=CF

=>CF=DB

b: Xét ΔBDC và ΔFCD có

BD=CF

DC chung

BC=DF

=>ΔBDC=ΔFCD

c: Xét ΔABC có AD/AB=AE/AC=1/2

nên DE//BC và DE=1/2BC