Câu 68 Trang 36: Tính
Lớp 9 SGK Toán tập 1

Câu 68 Trang 36: Tính

Ta có :

a.  $\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}=\sqrt[3]{3^{3}}-\sqrt[3]{(-2)^{3}}-\sqrt[3]{5^{3}}=3-(-2)-5=0$

b.  $\frac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}=\sqrt[3]{\frac{135}{5}}-\sqrt[3]{27}.\sqrt[3]{8}=3-3.2=-3$

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