Question

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)

(x+1)(x+2)(x+3)(x+4)=40

\(\left(x^2+x\right)4\left(x^2+x\right)=12\)

Answer 1

a: \(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

\(\Leftrightarrow x^4-14x^2+40-72=0\)

\(\Leftrightarrow\left(x^2-16\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow x\in\left\{4;-4\right\}\)

c: \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

\(\Leftrightarrow\left(x^2+x\right)^2+6\left(x^2+x\right)-2\left(x^2+x\right)-12=0\)

\(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)

=>(x+2)(x-1)=0

=>x=1 hoặc x=-2