Đặt x -2006 = y 

pt <=>  \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)

<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)

<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)

<=> \(49y^2-49y+49=57y^2-57y+19\)

<=> \(8y^2-8y-30=0\)

<=> \(4y^2-4y+15=0\)

Giải tiếp nha 

Ta có:\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=0\)

\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(loai\right)\\a=b=c\end{cases}}\)

\(\Rightarrow P=2007.2007.2007=2007^3\)

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)

\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)

\(\Leftrightarrow a\left(ab+ac+bc\right)+\left(b+c\right)\left(ab+ac+bc\right)-abc=0\)

\(\Leftrightarrow a\left(ab+ac+bc-bc\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)

\(\Leftrightarrow a^2\left(b+c\right)+\left(b+c\right)\left(ab+ac+bc\right)=0\)

\(\Leftrightarrow\left(a^2+ab+ac+bc\right)\left(b+c\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-c\\a=-b\\b=-c\end{matrix}\right.\)

- Nếu \(a=-c\Rightarrow a^{2006}=c^{2006}\Rightarrow c^{2006}-a^{2006}=0\Rightarrow P=0\)

- Nếu \(a=-b\Rightarrow a^{2004}=b^{2004}\Rightarrow a^{2004}-b^{2004}=0\Rightarrow P=0\)

- Nếu \(b=-c\Rightarrow b^{2005}=-c^{2005}\Rightarrow b^{2005}+c^{2005}=0\Rightarrow P=0\)

Vậy \(P=0\)

Đặt \(a=\sqrt{2006}-\sqrt{2005};b=\sqrt{2006}+\sqrt{2005}\)

Ta có 

\(a=\sqrt{2006}-\sqrt{2005}=\dfrac{\left(\sqrt{2006}-\sqrt{2005}\right)\left(\sqrt{2006}+\sqrt{2005}\right)}{\sqrt{2006}+\sqrt{2005}}=\dfrac{1}{b}\)

\(\RightarrowĐfcm\)

Hai bài này áp dụng hằng đẳng thức \(a^2-b^2=\left(a-b\right)\left(a+b\right)\) bạn nhé

a)

\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)\)

\(=2^2-\sqrt{3}^2\)

\(=4-3\)

\(=1\)

b)

Hai số nghịch đảo nhau là 2 số có tích của chúng bằng 1

Ví dụ

\(\frac{a}{b}\) và \(\frac{b}{a}\) ( hai số nghịch đảo )

\(\frac{a}{b}.\frac{b}{a}=1\)

Ta có

\(\left(\sqrt{2006}-\sqrt{2005}\right)\left(\sqrt{2006}+\sqrt{2005}\right)\)

\(=\sqrt{2006}^2-\sqrt{2005}^2\)

\(=2006-2005\)

\(=1\)

=> Đpcm 

mơn pn nhìu