View attachment 175834
C Đà Nẵng 2020
Em cảm ơn ạ.
đk: [tex]x^3\geq 7[/tex]
[tex]\Leftrightarrow 4x^3-x^2+2x-32+(x^3-4)(\sqrt{x^3-7}-1)=0[/tex]
[tex]\Leftrightarrow (x-2)(4x^2+7x+16)+(x^3-4)\frac{(x-2)(x^2+2x+4)}{\sqrt{x^3-7}+x^3-4}=0[/tex]
[tex]\Leftrightarrow (x-2)[(4x^2+7x+16)+\frac{(x^3-4)(x^2+2x+4)}{\sqrt{x^3-7}+x^3-4}]=0[/tex]
Do [tex](4x^2+7x+16)+\frac{(x^3-4)(x^2+2x+4)}{\sqrt{x^3-7}+x^3-4}>0;\forall x^3\geq 7[/tex]
[tex]\Rightarrow x=2[/tex]