T6

Question

Solve the equation:

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

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

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

$x=0$ .

U make by english, please – thanhnhan97gl 25-01-13 06:19 PM

2 Đáp án