Question

\(x^2+x-4+\left(x+6\right).\sqrt{x+2}=0\)

toán truong chuyen lam duoc tick cho

Answer 1

\(x^2+x-4+\left(x+6\right).\sqrt{x+2}=0\)

\(x^2+x-4+\left(x+6\right)\sqrt{x+2}=0\)

\(\left(x+6\right)\sqrt{x+2}=-x^2-x+4\)

\(\left(x+6\right)^2\left(x+2\right)=\left(-x^2-x+4\right)^2\)

\(x^3+2x^2+12x^2+24x+36x+72=\left(-x^2-x+4\right)\left(-x^2-x+4\right)\)

\(x^3+14x^2+60x+72=-\left(-x^4-x^3+4x^2\right)-\left(-x^3-x^2+4x\right)-4x^2-4x+16\)

\(x^3+14x^2+60x+72=x^4+x^3-4x^2+x^3+x^2-4x-4x^2-4x+16\)

\(x^3+14x^2+60x+72=x^4+2x^3-7x^2-8x+16\)

\(x^3+14x^2+60x+72-x^4-2x^3+7x^2+8x-16=0\)

\(-x^3+21x^2+68x+56-x^4=0\)

Đến đây chịu chắc phương trình vô nghiệm .