\(y^2=x\left(x+1\right)\left(x+7\right)\left(x+8\right)\)

\(=\left(x^2+8x\right)\left(x^2+8x+7\right)\)

\(\Rightarrow4y^2=\left(2x^2+16x\right)\left(2x^2+16x+14\right)\)

\(=\left(2x^2+16x+7-7\right)\left(2x^2+16x+7+7\right)\)

\(=\left(2x^2+16x+7\right)^2-49\)

\(\Leftrightarrow\left(2x^2+16x+7\right)^2-4y^2=49\)

\(\Leftrightarrow\left(2x^2+16x+7-2y\right)\left(2x^2+16x+7+2y\right)=49=1.49=7.7\)

Xét các trường hợp và thu được các nghiệm là: \(\left(-3,0\right),\left(0,0\right)\).

Áp dụng bđt svacxo\(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\), ta có:

A = \(\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}=\frac{a^2}{ab+ac-a^2}+\frac{b^2}{bc+ab-b^2}+\frac{c^2}{ac+bc-c^2}\)

=> A \(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)-\left(a^2+b^2+c^2\right)}=\frac{\left(a+b+c\right)^2}{4\left(ab+bc+ac\right)-\left(a+b+c\right)^2}\)

=> A \(\ge\frac{\left(a+b+c\right)^2}{\frac{4\left(a+b+c\right)^2}{3}-\left(a+b+c\right)^2}\)(bđt: \(xy+yz+xz\le\frac{\left(x+y+z\right)^2}{3}\))

=> A \(\ge\frac{3\left(a+b+c\right)^2}{4\left(a+b+c\right)^2-3\left(a+b+c\right)^2}=\frac{3\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=3\)

( x + 2 )( x² - x + 1 ) - ( x³ + 2x² )

= x( x² - x + 1 ) + 2( x² - x + 1 ) - x³ - 2x²

= x³ - x² + x + 2x² - 2x + 2 - x³ - 2x²

= -x² - x + 2 = ( 1 - x )( x + 2 )