T6

Question

Solve the equation:
$3x^4-4x^3=1-\sqrt{(1+x^2)^3}$

hay nhỉ, bạn học ĐH ngoại ngữ à, thế này các bác vô giải phải dịch ốm nhỉ :p

score 4 · Answer 1

Solution 2:

Applying the AM-GM inequality, we have,
$\left(1+\dfrac{3}{2}x^2\right)+\left(1+\dfrac{3}{2}x^2\right)+1\ge3\sqrt[3]{\left(1+\dfrac{3}{2}x^2\right)^2}$
or $1+x^2\ge\sqrt[3]{\left(1+\dfrac{3}{2}x^2\right)^2}$ or $\sqrt{(1+x^2)^3}\ge1+\dfrac{3}{2}x^2            (1)$
On the other hand,
     $x^4+2x^4+x^2+\dfrac{1}{2}x^2\ge4\sqrt[4]{x^12}=4x^3$
or $3x^4+\dfrac{3}{2}x^2\ge4x^3                           (2)$
From $(1)$ and $(2)$, we get: $3x^4-4x^3\ge1-\sqrt{(1+x^2)^3}$
Equality holds iff $x=0$

score 4 · Answer 2

Bình chọn tăng 4 Bình chọn giảm

The equation is equivalent to:
$3x^4-4x^3=(1-\sqrt{1+x^2})(2+x^2+\sqrt{1+x^2})$
or $x^2(3x^2-4x)=\dfrac{-x^2(2+x^2+\sqrt{1+x^2})}{1+\sqrt{1+x^2}}$
or $x^2\left(3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}\right)=0$
On the other hand, we have:
$3x^2-4x+\dfrac{2+x^2+\sqrt{1+x^2}}{1+\sqrt{1+x^2}}=3(x-\dfrac{2}{3})^2+\dfrac{2+5x^2+(\sqrt{1+x^2}-1)^2}{6(1+\sqrt{1+x^2})}>0$
then $x=0$.

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