\(3\sqrt[3]{x-2,3}=-6\)
\(\Leftrightarrow\sqrt[3]{x-2,3}=-\dfrac{6}{3}\)
\(\Leftrightarrow\sqrt[3]{x-2,3}=-2\)
\(\Leftrightarrow x-2,3=\left(-2\right)^3\)
\(\Leftrightarrow x-2,3=-8\)
\(\Leftrightarrow x=-8+2,3\)
\(\Leftrightarrow x=-5,7\)
Vậy: x=-5,7
$x=\root(3)(22\sqrt(2+)25-\root(3)(22\sqrt(2))- 25)$
root(2x + 2, 3) + root(2x + 1, 3) = root(2x ^ 2, 3) + root(2x ^ 2 + 1, 3)
Cho `2x^3=3y^3=4z^3`
`CMR:(\root{3}{2x^2+3y^2+4z^2})/(\root{3}{2}+\root{3}{3}+\root{3}{4})=1`
Giúp!
root(x ^ 2 + 2, 3) = 3
`(\root[3]{26+15\sqrt{3}}-\sqrt{3})/(\sqrt{6-2\sqrt{5}-\sqrt{5})`
root(5x + 2, 3) = 3
Giải phương trình, x>0
\(\frac{\left(x^3+3x^2\sqrt{x^3-3x+6}\right)\left(3x-x^3-2\right)}{2+\sqrt{x^3-3x+6}}=4\left[2\sqrt{\left(x^3-3x+6\right)^3}-\left(x^3-3x+6\right)^2\right]\)
Giải phương trình, x>0
\(\frac{\left(x^3+3x^2\sqrt{x^3-3x+6}\right)\left(3x-x^3-2\right)}{2+\sqrt{x^3-3x+6}}=4\left[2\sqrt{\left(x^3-3x+6\right)^3}-\left(x^3-3x+6\right)^2\right]\)
Đặt $x=\sqrt[3]{3+2\sqrt{2}},y=\sqrt[3]{3-2\sqrt{2}}$
$\Rightarrow \left\{\begin{matrix} x^{3}+y^{3}=6\\xy=1 \end{matrix}\right.$
$\Rightarrow (x+y)^{3}=x^{3}+y^{3}+3xy(x+y)=6+3xy=3[1+1+(x+y)]> 3.3\sqrt[3]{1.1.(x+y)}$
(Vì x>1,y>0=>x+y>1)
Do đó: $(x+y)^{3}> 3^{2}.\sqrt[3]{x+y}$
$\Rightarrow (x+y)^{9}>3^{6}.(x+y)$
$\Rightarrow (x+y)^{8}>3^{6}$
=>đpcm
