Bài tập So sánh hai căn bậc ba lớp 9 (có lời giải)

Đáp án đúng là: B

Ta có: 3 = \(\sqrt[3]{{27}}\) < \(\sqrt[3]{{27\frac{1}{4}}}\) nên 3 < \(\sqrt[3]{{27\frac{1}{4}}}\).

Đáp án đúng là: A

Ta có: \(2\sqrt[3]{3} = \sqrt[3]{{8.3}} = \sqrt[3]{{24}} > \sqrt[3]{{23}}\) nên \(2\sqrt[3]{3} > \sqrt[3]{{23}}\).

Đáp án đúng là: A

Ta có: 12 = \(\sqrt[3]{{{{12}^3}}} = \sqrt[3]{{1728}} < \sqrt[3]{{1730}}\).

Do đó, 4.12 < 4. \(\sqrt[3]{{1730}}\) hay 48 < 4. \(\sqrt[3]{{1730}}\).

Đáp án đúng là: A

Ta có: \(\frac{2}{3}\sqrt[3]{{18}} = \sqrt[3]{{18.{{\left( {\frac{2}{3}} \right)}^3}}} = \sqrt[3]{{18.\frac{8}{{27}}}} = \sqrt[3]{{\frac{{16}}{3}}} = \sqrt[3]{{5\frac{1}{3}}}\).

\(\frac{3}{4}\sqrt[3]{{12}} = \sqrt[3]{{12.{{\left( {\frac{3}{4}} \right)}^3}}} = \sqrt[3]{{12.\frac{{27}}{{64}}}} = \sqrt[3]{{\frac{{81}}{{16}}}} = \sqrt[3]{{5\frac{1}{{16}}}}\).

Ta có: \(5\frac{1}{3} > 5\frac{1}{{16}}\) nên \(\frac{2}{3}\sqrt[3]{{18}} > \frac{3}{4}\sqrt[3]{{12}}\).

Đáp án đúng là: A

Ta có: \(3\sqrt[3]{{133}} = \sqrt[3]{{3591}}\) và 33 = \(\sqrt[3]{{35937}}\).

Vì \(\sqrt[3]{{35937}}\) > \(\sqrt[3]{{3591}}\) nên 33 > \(3\sqrt[3]{{133}}\).

Đáp án đúng là: A

Ta có: \(3\sqrt[3]{{12}} = \sqrt[3]{{{{12.3}^3}}} = \sqrt[3]{{324}}\).

Do đó, \(3\sqrt[3]{{12}} - 1 = \sqrt[3]{{324}} - 1 < \sqrt[3]{{343}} - 1 = 7 - 1 = 6\)

Có \(\sqrt[3]{{130}} + 1 > \sqrt[3]{{125}} + 1 = 5 + 1 = 6\).

Suy ra \(\sqrt[3]{{130}} + 1 > 3\sqrt[3]{{12}} - 1\).

Đáp án đúng là: B

Ta có: \(\sqrt[3]{{\frac{{ - 1000}}{{216}}}} = \sqrt[3]{{{{\left( {\frac{{ - 10}}{6}} \right)}^3}}} = - \frac{{10}}{6}\);

\(\sqrt[3]{{\frac{{ - 729}}{{125}}}} = \sqrt[3]{{{{\left( {\frac{{ - 9}}{5}} \right)}^3}}} = - \frac{9}{5}\).

Ta có: \( - \frac{{10}}{6} = \frac{{ - 50}}{{30}} < \frac{{ - 36}}{{30}} = - \frac{9}{5}\).

Vậy \(\sqrt[3]{{\frac{{ - 1000}}{{216}}}} < \sqrt[3]{{\frac{{ - 729}}{{125}}}}\).

Đáp án đúng là: A

Ta có: \(\sqrt[3]{{\frac{1}{{125}}}} + \sqrt[3]{{ - \frac{1}{{729}}}} = - \frac{1}{5} + \left( { - \frac{1}{9}} \right) = - \frac{{14}}{{45}}\);

\(\sqrt[3]{{\frac{1}{{216}}}} + \sqrt[3]{{ - \frac{1}{{512}}}} = \frac{1}{6} - \frac{1}{5} = - \frac{1}{{30}}\).

Do \( - \frac{1}{{30}} = \frac{{ - 3}}{{90}} > \frac{{ - 28}}{{90}} = \frac{{ - 14}}{{45}}\) nên \(\sqrt[3]{{\frac{1}{{125}}}} + \sqrt[3]{{ - \frac{1}{{729}}}} > \sqrt[3]{{\frac{1}{{216}}}} + \sqrt[3]{{ - \frac{1}{{512}}}}\).