Ta có: \(x\left(x+2\right)+x-2=0\)

\(\Leftrightarrow x^2+2x+x-2=0\)

\(\Leftrightarrow\left(x^2+3x+\frac{9}{4}\right)-\frac{17}{4}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\left(\frac{\sqrt{17}}{2}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{3-\sqrt{17}}{2}\right)\left(x+\frac{3+\sqrt{17}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-3+\sqrt{17}}{2}\\x=-\frac{3+\sqrt{17}}{2}\end{cases}}\)

khó thế nhờ (^o^)

P = 2a² + 9a - 6

= 2( a² + 9/2a + 81/16 ) - 129/8

= 2( a + 9/4 )² - 129/8 ≥ -129/8 ∀ a

Dấu "=" xảy ra khi a = -9/4

=> MinP = -129/8 <=> a = -9/4

\(\text{KQ quen thuộc:}xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+2xyz=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

do đó: \(x=-y\text{ hoặc: }y=-z\text{ hoặc: }z=-x\) do đó: A=0

phân tích vế trái thành nhân tử sau đó tính A

           Bài làm :

Ta có :

\(\left(a^2+2017\right)\left(b^2+2017\right)\left(c^2+2017\right)\)

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

\(=\left[\left(a^2+ab\right)+\left(bc+ca\right)\right]\left[\left(b^2+ab\right)+\left(bc+ca\right)\right]\left[\left(c^2+bc\right)+\left(ab+ca\right)\right]\)

\(=\left[a\left(a+b\right)+c\left(b+a\right)\right]\left[b\left(b+a\right)+c\left(b+a\right)\right]\left[c\left(c+b\right)+a\left(b+c\right)\right]\)\(=\left(a+b\right)\left(c+a\right)\left(a+b\right)\left(b+c\right)\left(b+c\right)\left(c+a\right)\)

\(=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

=> Điều phải chứng minh

             Bài làm :

Ta có :

\(...\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2-4=0\)

\(\Leftrightarrow12x+4=0\)

\(\Leftrightarrow12x=-4\)

\(\Leftrightarrow x=-\frac{1}{3}\)