12 bài tập Rút gọn biểu thức chứa căn bậc ba có lời giải

Đáp án đúng là: A

$\frac{{\sqrt[3]{2}}}{{\sqrt[3]{2} - 1}} - \sqrt[3]{4} - \sqrt[3]{2} = \frac{{\sqrt[3]{2}\left( {\sqrt[3]{{{2^2}}} + \sqrt[3]{2} + 1} \right)}}{{\left( {\sqrt[3]{2} - 1} \right)\left( {\sqrt[3]{{{2^2}}} + \sqrt[3]{2} + 1} \right)}} - \sqrt[3]{4} - \sqrt[3]{2}$

$= 2 + \sqrt[3]{4} + \sqrt[3]{2} - \sqrt[3]{4} - \sqrt[3]{2} = 2$.

Đáp án đúng là: A

Ta có: $\sqrt[3]{{\left( {\sqrt 2 + 1} \right)\left( {3 + 2\sqrt 2 } \right)}} = \sqrt[3]{{\left( {\sqrt 2 + 1} \right){{\left( {1 + \sqrt 2 } \right)}^2}}} = \sqrt[3]{{{{\left( {\sqrt 2 + 1} \right)}^3}}} = \sqrt 2 + 1$.

Đáp án đúng là: A

Ta có: $\sqrt[3]{{ - 343}}.\sqrt[3]{3} + \sqrt[3]{{81}} - 2\sqrt[3]{{24}} = - 7\sqrt[3]{3} + 3.3\sqrt[3]{3} - 2\sqrt[3]{3} = 0$.

Đáp án đúng là: C

Ta có:

$2\sqrt[3]{{24}} - 3\sqrt[3]{{81}} + 4\sqrt[3]{{192}} - 2\sqrt[3]{{375}}$

$= 2\sqrt[3]{{8.3}} - 3\sqrt[3]{{27.3}} + 4\sqrt[3]{{64.3}} - 2\sqrt[3]{{125.3}}$

$= 2.2\sqrt[3]{3} - 3.3\sqrt[3]{3} + 4.4\sqrt[3]{3} - 2.5.\sqrt[3]{3}$

$= 4\sqrt[3]{3} - 9\sqrt[3]{3} + 16\sqrt[3]{3} - 10.\sqrt[3]{3} = \sqrt[3]{3}$.

Đáp án đúng là: D

Ta có:

$\left( {\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}} \right)\left( {\sqrt[3]{3} + \sqrt[3]{2}} \right)$

$= \sqrt[3]{9}.\sqrt[3]{3} - \sqrt[3]{6}.\sqrt[3]{3} + \sqrt[3]{4}.\sqrt[3]{3} + \sqrt[3]{9}.\sqrt[3]{2} - \sqrt[3]{6}.\sqrt[3]{2} + \sqrt[3]{4}.\sqrt[3]{2}$

$= \sqrt[3]{{{3^3}}} + \sqrt[3]{{18}} - \sqrt[3]{{18}} - \sqrt[3]{{12}} + \sqrt[3]{{12}} + \sqrt[3]{{{2^3}}}$

$= 3 + 2 = 5$.

Đáp án đúng là: A

Ta có: A = $\sqrt[3]{{\frac{{343{a^3}{b^6}}}{{ - 216}}}} = - \frac{7}{6}a{b^2}$.

Đáp án đúng là: C

Ta có: B = $3x - \sqrt[3]{{27{x^3} + 27{x^2} + 9x + 1}}$

$3x - \sqrt[3]{{{{\left( {3x + 1} \right)}^3}}} = 3x - 3x - 1 = - 1$.

Đáp án đúng là: B

Ta có: C = $2\sqrt[3]{{1 - 3{x^2} + 3x - {x^3}}} - x$

= $2\sqrt[3]{{{{\left( {1 - x} \right)}^3}}} - x = 2\left( {1 - x} \right) - x = 2 - 3x$.

Đáp án đúng là: D

Ta có: D = $\sqrt[3]{{{{\left( {x - 1} \right)}^3}}} - \frac{1}{2}\sqrt[3]{{{{\left( {1 - 2x} \right)}^3}}} + 3x$

= $x - 1 - \frac{1}{2}\left( {1 - 2x} \right) + 3x = x - 1 - \frac{1}{2} + x + 3x = 5x - \frac{3}{2}$.

Đáp án đúng là: C

Ta có:

E = $\sqrt[3]{{x\sqrt x + 1}}.\sqrt[3]{{x\sqrt x - 1}} - \sqrt[3]{{1 - {x^3}}}$

= $\sqrt[3]{{\left( {x\sqrt x + 1} \right)\left( {x\sqrt x - 1} \right)}} - \sqrt[3]{{1 - {x^3}}}$

= $\sqrt[3]{{{x^3} - 1}} - \sqrt[3]{{1 - {x^3}}} = 2\sqrt[3]{{{x^3} - 1}}$.

a) $\frac{{\sqrt[3]{{108}}}}{{\sqrt[3]{4}}} + \frac{{\sqrt[3]{{7{,}2}}}}{{\sqrt[3]{{0{,}9}}}} = \sqrt[3]{{\frac{{108}}{4}}} + \sqrt[3]{{\frac{{7{,}2}}{{0{,}9}}}} = \sqrt[3]{{27}} + \sqrt[3]{8} = 5$.

b) $2\sqrt[3]{{24}} - 5\sqrt[3]{{81}} + 4\sqrt[3]{{192}} = 2.2\sqrt[3]{3} - 5.3.\sqrt[3]{3} + 4.4.\sqrt[3]{3} = 5\sqrt[3]{3}$.

a) $\sqrt[3]{{64{a^3}}} = \sqrt[3]{{{{\left( {8a} \right)}^3}}} = 8a$;

b) $\sqrt[3]{{ - 8{a^3}{b^6}}} = \sqrt[3]{{{{\left( { - 2a{b^2}} \right)}^3}}} = - 2a{b^2}$.