Đáp án+giải thích các bước giải:
1, so sán:
Ta có:
$5\sqrt[3]{6}=\sqrt[3]{125}.\sqrt[3]{6}=\sqrt[3]{125.6}=\sqrt[3]{750}$
$6\sqrt[3]{5}=\sqrt[3]{216}.\sqrt[3]{5}=\sqrt[3]{216.5}=\sqrt[3]{1080}$
do $\sqrt[3]{750}<\sqrt[3]{1080}$
nên ⇒ $5\sqrt[3]{6}<6\sqrt[3]{5}$
2, rút gọn:
$\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}:\sqrt[3]{4}$
$=\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\dfrac{\sqrt[3]{54}}{\sqrt[3]{4}}$
$=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{\dfrac{54}{4}}$
$=\sqrt[3]{27}-\sqrt[3]{\dfrac{27}{2}}$
$=3-\dfrac{\sqrt[3]{27}}{\sqrt[3]{2}}$
$=3-\dfrac{3}{\sqrt[3]{2}}$
3, tìm x:
a)
$\sqrt[3]{2x+1}=3$
$\sqrt[3]{2x+1}=\sqrt[3]{27}$
$2x+1=27$
$2x=26$
$x=13$
$\text{Vậy a=13}$
b)
$\sqrt[3]{x-1} +1 =x$
$\sqrt[3]{x-1}=x-1$
$\sqrt[3]{x-1}=\sqrt[3]{(x-1)^3}$
$x-1=(x-1)^3$
$(x-1)-(x-1)^3=0$
$(x-1)[1-(x-1)^2)=0$
$(x-1)(1-x+1)(1+x-1)=0$
$x(x-1)(2-x)=0$
⇒\(\left[ \begin{array}{l}x=0\\x-1=0\\2-x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=1\\x=2\end{array} \right.\)
$\text{Vậy...}$
