Question

Giải phương trình

\(\left(x^2+x+1\right)^2=3\left(x^4+x^2+1\right)\)

\(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

\(2x\left(8x-1\right)^2\left(4x-1\right)=9\)

\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)

Answer 1

Answer 2

a)\(3\left(x^4+x^2+1\right)=\left(x^2+x+1\right)^2\)

Cauchy-schwarz:

\(\left(1+1+1\right)\left(x^4+x^2+1\right)\ge\left(x^2+x+1\right)^2\)

"="<=>\(x=1\)

b)\(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

\(x^2+x-1=t\)

\(\Rightarrow\left(t-1\right)\left(t+1\right)=24\)

\(\Leftrightarrow t^2-25=0\)

\(\Leftrightarrow t=\pm5\)

t=5\(\Leftrightarrow x^2+x-1=5\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

t=-5<=> pt vô nghiệm

Answer 3

Answer 4

Answer 5

Answer 6