Bài 68 :

a ) \(\sqrt[3]{27}-\sqrt[3]{8}-\sqrt[3]{125}=3-2-5=-4\)

b ) \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54.4}=\sqrt[3]{27}-\sqrt[3]{216}=3-6=-3\)

Bài 69 :

a ) Ta có : \(\left\{{}\begin{matrix}3^3=27\\\left(\sqrt[3]{123}\right)^3=123\end{matrix}\right.\)

Vì 27 < 123 nên suy ra \(3< \sqrt[3]{123}\)

Vậy \(3< \sqrt[3]{123}\)

a) \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)

= \(\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)

= \(-16\sqrt{3}\)

b) \(\left(a.\sqrt{\dfrac{a}{b}}+2\sqrt{ab}+b.\sqrt{\dfrac{b}{a}}\right)\sqrt{\dfrac{a}{b}}\)

= \(\dfrac{a^2}{b}+2a+b\) = \(\dfrac{a^2+\left(2a+b\right)b}{b}\) = \(\dfrac{a^2+2ab+b^2}{b}\) = \(\dfrac{\left(a+b\right)^2}{b}\)

c) \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\) = \(3+2-5=0\)

d) \(3+\sqrt{18}+\sqrt{3}+\sqrt{8}\) = \(3+3\sqrt{2}+\sqrt{3}+2\sqrt{2}\)

= \(3+\sqrt{3}+5\sqrt{2}\)

LG a

3√27−3√−8−3√125273−−83−1253

Phương pháp giải:

Tính từng căn bậc ba rồi thực hiện phép tính

Lời giải chi tiết:

3√27−3√−8−3√125=3√33−3√(−2)3−3√53273−−83−1253=333−(−2)33−533

=3−(−2)−5=3−(−2)−5

=3+2−5=0=3+2−5=0.

LG b

 3√1353√5−3√54.3√4135353−543.43

Phương pháp giải:

Sử dụng các công thức:

3√a.b=3√a.3√ba.b3=a3.b3.

3√ab=3√a3√bab3=a3b3,  với b≠0b≠0.

Lời giải chi tiết:

3√1353√5−3√54.3√4=3√27.53√5−3√54.4135353−543.43=27.5353−54.43

=3√5.3√273√5−3√216=53.27353−2163

=3√27−3√216=273−2163

=3√33−3√63=333−633

=3−6=−3=3−6=−3.

a) \sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}=3-(-2)-5=0327​−3−8​−3125​=3−(−2)−5=0.

b) \dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54} \cdot \sqrt[3]{4}=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54.4}=3-6=-335​3135​​−354​⋅34​=35135​​−354.4​=3−6=−3.

a) \sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}=3-(-2)-5=0327​−3−8​−3125​=3−(−2)−5=0.

b) \dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54} \cdot \sqrt[3]{4}=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54.4}=3-6=-335​3135​​−354​⋅34​=35135​​−354.4​=3−6=−3.

b: Ta có: \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}\cdot\sqrt[3]{4}\)

\(=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54\cdot4}\)

=3-6

=-3

a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)

a)\(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\)

\(=3+2-5\)

\(=0\)

b)\(\frac{\sqrt[3]{153}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)

\(=\sqrt[3]{\frac{153}{5}}-\sqrt[3]{54.4}\)

\(=\sqrt[3]{\frac{153}{5}}-6\)

Theo mình câu b như vậy

pham trung thanh câu b bn làm thiếu hay sao ý? Theo tôi, cả bài làm như thế này.

Giải:

a, \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\)

\(=\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{12}=3+2-5\)

\(=0\)

b, \(\frac{\sqrt[3]{153}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)

\(=\sqrt[3]{\frac{135}{5}}-\sqrt[3]{54.4}\)

\(=\sqrt[3]{27}-\sqrt[3]{216}\)

\(=3-6\)

\(=-3\)

Bạn ghi sai đề, bạn ghi là 153, nếu là 135 thì tôi lại giải ra đc luôn, bạn nhìn lại đề xem

`a)\root[3]{135}/\root[3]{5}-\root[3]{54}.\root[3]{4}`

`=\root[3]{135/5}-\root[3]{54.4}`

`=\root[3]{27}-\root[3]{216}`

`=3-6=-3`

`b)(\root[3]{25}-\root[3]{10}+\root[3]{4})(\root[3]{5}+\root[3]{2})`

`=5+\root[3]{50}-\root[3]{50}-\root[3]{20}+\root[3]{20}+2`

`=7`.

a: Ta có: \(A=\sqrt{8}-2\sqrt{18}+3\sqrt{50}\)

\(=2\sqrt{2}-6\sqrt{2}+15\sqrt{2}\)

\(=11\sqrt{2}\)

b: Ta có: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)

\(=5\sqrt{5}-\sqrt{5}+\sqrt{5}-1\)

\(=5\sqrt{5}-1\)