10 bài tập Toán 9 Kết nối tri thức Bài 10. Căn bậc ba và căn thức bậc ba có đáp án

a) $\sqrt[3]{{64}} = 4$.

b) $\sqrt[3]{{ - 512}} = - 8$

c) $\sqrt[3]{{0{,}064}} = \sqrt[3]{{\frac{{64}}{{1000}}}} = \frac{{\sqrt[3]{{64}}}}{{\sqrt[3]{{1000}}}} = \frac{4}{{10}}$.

d) $\sqrt[3]{{ - 0{,}216}} = \sqrt[3]{{{{\left( { - 6} \right)}^3}}} = - 6$.

e) $\frac{{\sqrt[3]{{500}}}}{{\sqrt[3]{4}}} + \sqrt[3]{{12}} \cdot \sqrt[3]{{18}} = \sqrt[3]{{\frac{{500}}{4}}} + \sqrt[3]{{12 \cdot 18}} = \sqrt[3]{{125}} + \sqrt[3]{{216}} = 5 + 6 = 11$

f) $\frac{{\sqrt[3]{{12}} \cdot \sqrt[3]{6}}}{{\sqrt[3]{{576}}}} - \frac{{\sqrt[3]{{32}}}}{{\sqrt[3]{4}}} = \sqrt[3]{{\frac{{12 \cdot 6}}{{576}}}} - \sqrt[3]{{\frac{{32}}{4}}} = \sqrt[3]{{\frac{1}{8}}} - \sqrt[3]{8} = \frac{1}{2} - 2 = - \frac{3}{2}$.

a) $\sqrt[3]{{\frac{1}{2}}} \cdot \sqrt[3]{{ - 18}} \cdot \sqrt[3]{3} = \sqrt[3]{{\frac{1}{2} \cdot ( - 18) \cdot 3}} = \sqrt[3]{{ - 27}} = - 3$.

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

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

c) $\left( {\frac{1}{2}\sqrt[3]{2} - \frac{1}{4}\sqrt[3]{{16}}} \right) \cdot \sqrt[3]{4} = \left( {\frac{1}{2}\sqrt[3]{2} - \frac{1}{2}\sqrt[3]{2}} \right) \cdot \sqrt[3]{4} = 0$.

d) $\left( {\frac{1}{2}\sqrt[3]{9} - 2\sqrt[3]{3} + 3\sqrt[3]{{\frac{1}{3}}}} \right):2\sqrt[3]{{\frac{1}{3}}} = \left( {\frac{1}{2}\sqrt[3]{9} - 2\sqrt[3]{3} + \sqrt[3]{9}} \right):\frac{2}{3}\sqrt[3]{9} = \left( {\frac{3}{2}\sqrt[3]{9} - 2\sqrt[3]{3}} \right):\frac{2}{3}\sqrt[3]{9} = \frac{9}{4} - \sqrt[3]{9}$.

e) $\left( {\sqrt[3]{9} - \sqrt[3]{6} + \sqrt[3]{4}} \right)\left( {\sqrt[3]{3} + \sqrt[3]{2}} \right) = \left( {\sqrt[3]{{{3^2}}} - \sqrt[3]{{3 \cdot 2}} + \sqrt[3]{{{2^2}}}} \right)\left( {\sqrt[3]{3} + \sqrt[3]{2}} \right) = {\left( {\sqrt[3]{3}} \right)^3} + {\left( {\sqrt[3]{2}} \right)^3} = 3 + 2 = 5$.

a) $\sqrt[3]{{ - 64}} - \sqrt[3]{{125}} + \sqrt[3]{{216}} = - 4 - 5 + 6 = - 3$

b) Ta có ${\left( {\sqrt[3]{4} + 1} \right)^3} - {\left( {\sqrt[3]{4} - 1} \right)^3}$

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

$$= 2.\left( {\sqrt[3]{{16}} + 2\sqrt[3]{4} + 1 + \sqrt[3]{{16}} - 1 + \sqrt[3]{{16}} - 2\sqrt[3]{4} + 1} \right) = 2\left( {3\sqrt[3]{{16}} + 1} \right) = 2\left( {6\sqrt[3]{2} + 1} \right)$$

c) Ta có $\left( {12\sqrt[3]{2} + \sqrt[3]{{16}} - 2\sqrt[3]{2}} \right)\left( {5\sqrt[3]{4} - 3\sqrt[3]{{\frac{1}{2}}}} \right)$

$$= \left( {12\sqrt[3]{2} + 2\sqrt[3]{2} - 2\sqrt[3]{2}} \right)\left( {5\sqrt[3]{4} - 3\sqrt[3]{{\frac{1}{2}}}} \right) = 12\sqrt[3]{2}\left( {5\sqrt[3]{4} - 3\sqrt[3]{{\frac{1}{2}}}} \right) = 120 - 36 = 84$$

Ta có $x = \frac{2}{{2\sqrt[3]{2} + 2 + \sqrt[3]{4}}} = \frac{2}{{\sqrt[3]{{16}} + 2 + \sqrt[3]{4}}}$

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

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

Tương tự $y = \frac{6}{{2\sqrt[3]{2} - 2 + \sqrt[3]{4}}} = \sqrt[3]{4} + \sqrt[3]{2}$

Do đó: $x{y^3} - {x^3}y = xy\left( {{y^2} - {x^2}} \right) = xy\left( {y - x} \right)\left( {y + x} \right) = 8\left( {2\sqrt[3]{2} - \sqrt[3]{4}} \right)$

a) $A = \sqrt[3]{{{2^3} + 3 \cdot {2^2}\left( {\sqrt 2 } \right) + 3 \cdot 2 \cdot {{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^3}}} + \sqrt[3]{{{2^3} - 3 \cdot {2^2}\left( {\sqrt 2 } \right) + 3 \cdot 2 \cdot {{\left( {\sqrt 2 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^3}}}$

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

b) ${B^3} = 182 + \sqrt {33125} + 182 - \sqrt {33125} + 3\sqrt[3]{{{{182}^2} - 33125}}B = 364 - 3B.$

Khi đó ${B^3} + 3B - 364 = 0$ nên $\left( {B - 7} \right)\left( {{B^2} + 7B + 52} \right) = 0$, suy ra $B = 7$.

(do ${B^2} + 7B + 52 = {\left( {B + \frac{7}{2}} \right)^2} + \frac{{159}}{4} > 0$).

a) Ta có $A = \left[ {\left( {\frac{1}{a} - \sqrt[6]{{\frac{1}{a}}} + \sqrt[3]{{{a^2}}}} \right) + \left( {\frac{a}{{{a^2}}}\sqrt[6]{{{a^5}}} - \frac{3}{a}\sqrt[3]{{{a^2}}}} \right)} \right]a\sqrt[3]{a}$

$= \left[ {\left( {\frac{1}{a} - \sqrt[6]{{\frac{1}{a}}} + \sqrt[3]{{{a^2}}}} \right) + \left( {\sqrt[6]{{\frac{1}{a}}} - 3\sqrt[3]{{\frac{1}{a}}}} \right)} \right]a\sqrt[3]{a}$

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

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

b) Ta có $B = \sqrt[3]{{{a^2}}}\sqrt {\sqrt[3]{{{a^2}}} + \sqrt[3]{{{b^2}}}} + \sqrt[3]{{{b^2}}}\sqrt {\sqrt[3]{{{a^2}}} + \sqrt[3]{{{b^2}}}}$

$= \left( {\sqrt[3]{{{a^2}}} + \sqrt[3]{{{b^2}}}} \right)\sqrt {\sqrt[3]{{{a^2}}} + \sqrt[3]{{{b^2}}}}$

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

a) $\sqrt[3]{{{x^3} + 9{x^2}}} = x + 3$

$\sqrt[3]{{{x^3} + 9{x^2}}} = \sqrt[3]{{{{\left( {x + 3} \right)}^3}}}$

$\begin{array}{l}{x^3} + 9{x^2} = {\left( {x + 3} \right)^3} = {x^3} + 9{x^2} + 27x + 27\\27x + 27 = 0\\x = - 1.\end{array}$

Vậy $x = - 1.$

b) $\sqrt[3]{{5 + x}} - x = 5$

$$\begin{array}{l}\sqrt[3]{{5 + x}} = x + 5\\x + 5 = {\left( {x + 5} \right)^3}\\\left( {x + 5} \right)\left[ {{{\left( {x + 5} \right)}^2} - 1} \right] = 0\end{array}$$

TH1: $x + 5 = 0$ nên $x = - 5$.

TH1: ${\left( {x + 5} \right)^2} = 1$

$x + 5 = 1$ hoặc $x + 5 = - 1$

$x = - 4$ hoặc $x = - 6$

Vậy $x \in \left\{ { - 5\,;\,\,4\,;\,\, - 6} \right\}.$

a) $\sqrt[3]{{1\,\,000x}} - \sqrt[3]{{64x}} - \sqrt[3]{{27x}} = 15$ $\begin{array}{l}10\sqrt[3]{x} - 4\sqrt[3]{x} - 3\sqrt[3]{x} = 15\\\sqrt[3]{x} = 5\\x = 125\end{array}$

Vậy phương trình có một nghiệm $x = 125$.

b) $\sqrt[3]{{x - 3}} + 3 = x$

$\begin{array}{l}\sqrt[3]{{x - 3}} = x - 3\\x - 3 = {\left( {x - 3} \right)^3}\end{array}$

$\left( {x - 3} \right)\left[ {{{\left( {x - 3} \right)}^2} - 1} \right] = 0$

$\left( {x - 3} \right)\left( {x - 2} \right)\left( {x - 4} \right) = 0$

$x = 3$ hoặc $x = 2$ hoặc $x = 4$

Vậy tập nghiệm của phương trình là $S = \{ 2;3;4\}$.

a) $\sqrt[3]{{2x - 1}} + \sqrt[3]{{x - 1}} = \sqrt[3]{{3x - 2}}$

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

$3\sqrt[3]{{\left( {2x - 1} \right)\left( {x - 1} \right)\left( {3x - 2} \right)}} = 0$

$x = \frac{1}{2}$ hoặc $x = 1$ hoặc $x = \frac{2}{3}$.

Thử lại ta thấy các nghiệm đều thỏa mãn phương trình.

Vậy phương trình đã cho có ba nghiệm là $x = \frac{1}{2}$; $x = 1$ và $x = \frac{2}{3}$.

b) $\sqrt[3]{{x + 5}} + \sqrt[3]{{x + 6}} = \sqrt[3]{{2x + 11}}$

$x + 5 + x + 6 + 3\sqrt[3]{{\left( {x + 5} \right)\left( {x + 6} \right)}}\left( {\sqrt[3]{{x + 5}} + \sqrt[3]{{x + 6}}} \right) = 2x + 11$

$3\sqrt[3]{{\left( {x + 5} \right)\left( {x + 6} \right)\left( {2x + 11} \right)}} = 0$

$x = - 5$ hoặc $x = - 6$ hoặc $x = - \frac{{11}}{2}.$

Thử lại ta thấy các nghiệm đều thỏa mãn phương trình.

Vậy tập nghiệm của phương trình là $S = \left\{ { - 5; - \frac{{11}}{2}; - 6} \right\}$.