Question

1.Tìm x :

\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=-15\)

2.Cho a,b,c khác 0 va a+b+c=abc ; \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

C/m \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)

Answer 1

1 ) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=-15\)

\(\Leftrightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=-15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=-15\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(6x^2-6x^2\right)+\left(12x+12x\right)+\left(27+6-8\right)=-15\)

\(\Leftrightarrow24x+25=-15\)

\(\Leftrightarrow24x=-40\)

\(\Leftrightarrow x=-\dfrac{5}{3}\)

Vậy \(x=-\dfrac{5}{3}\)