Đáp án:

Giải thích các bước giải:

6.8:
Ta có: $(a+b+c)^2=a^2+b^2+c^2$

⇔ $a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2$

⇔ $2(ab+bc+ac)=0$

⇔ $ab+bc+ac=0$
Ta lại có:
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

$=\dfrac{ab+bc+ac}{abc}$

$=0$
Mặt khác:
$\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}$

$=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}$

$=abc(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3})$

Cần chứng minh:
$\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}$

Ta có:
$\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}$

$=(\dfrac{1}{a}+\dfrac{1}{b})^3-\dfrac{3}{ab}.(\dfrac{1}{a}+\dfrac{1}{b})+\dfrac{1}{c^3}$

$=(\dfrac{1}{a}+\dfrac{1}{b})^3+\dfrac{1}{c^3}-\dfrac{3}{ab}.(\dfrac{1}{a}+\dfrac{1}{b})$

$=(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})[(\dfrac{1}{a}+\dfrac{1}{b})^2-\dfrac{1}{c}.(\dfrac{1}{a}+\dfrac{1}{b})]+\dfrac{1}{c^2}-\dfrac{3}{ab}.(\dfrac{1}{a}+\dfrac{1}{b})$

$=0-\dfrac{3}{ab}.(\dfrac{1}{a}+\dfrac{1}{b})$

$=-\dfrac{3}{ab}.(-\dfrac{1}{c})$

$=\dfrac{3}{abc}$
⇒ đpcm

6.9:
$E=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}$

⇔ $E=\dfrac{ac}{abc+ac+c}+\dfrac{abc}{abc^2+abc+ac}+\dfrac{c}{ac+c+1}$

⇔ $E=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}$

⇔ $E=\dfrac{ac+c+1}{ac+c+1}$

⇔ $E=1$
Chúc bạn học tốt !!