Bài 9: Căn bậc ba

a) \(3\sqrt[3]{2}=\sqrt[3]{3^3.2}=\sqrt[3]{54}>\sqrt[3]{53}\)

b) \(22=\sqrt[3]{22^3}=\sqrt[3]{10648}>\sqrt[3]{10638}=3\sqrt[3]{394}\)

a) \(6=\sqrt[3]{6^3}=\sqrt{216}>\sqrt[3]{208}=2\sqrt[3]{26}\)

b) \(2\sqrt[3]{6}=\sqrt[3]{2^3.6}=\sqrt[3]{48}>\sqrt[3]{47}\)

a: Sửa đề: \(\sqrt[3]{\left(4-2\sqrt{3}\right)\cdot\left(\sqrt{3}-1\right)}\)

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

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

b: \(\sqrt{3+\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}}\)

\(=\sqrt{3+\sqrt{3}+\sqrt[3]{\left(\sqrt{3}\right)^3+3\cdot\left(\sqrt{3}\right)^2\cdot1+3\cdot\sqrt{3}\cdot1^2+1^3}}\)

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

\(=\sqrt{3+\sqrt{3}+\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

a: Sửa đề: \(M=3x-\sqrt[3]{27x^3+27x^2+9x+1}\)

\(=3x-\sqrt[3]{\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3}\)

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

\(=3x-3x-1=-1\)

b: \(N=\sqrt[3]{8x^3+12x^2+6x+1}-\sqrt[3]{x^3}\)

\(=\sqrt[3]{\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3}-x\)

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

=2x+1-x

=x+1

`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`.

\(M=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\)

=>\(M^3=7+5\sqrt{2}+7-5\sqrt{2}+3\cdot M\cdot\sqrt[3]{\left(7+5\sqrt{2}\right)\left(7-5\sqrt{2}\right)}\)

=>\(M^3=14+3M\cdot\left(-1\right)=14-3M\)

=>\(M^3+3M-14=0\)

=>\(M^3-2M^2+2M^2-4M+7M-14=0\)

=>\(\left(M-2\right)\left(M^2+2M+7\right)=0\)

=>M-2=0

=>M=2

\(\Leftrightarrow M=\dfrac{4}{2}=\dfrac{4}{\sqrt[3]{8}}>\dfrac{4}{\sqrt[3]{9}}=N\)

\(\left(\sqrt[3]{\dfrac{1}{9}}+4\cdot\sqrt[3]{\dfrac{1}{72}}-\sqrt[3]{4}\right)\left(\sqrt[3]{72}+\sqrt[3]{96}+\sqrt[3]{128}\right)\)

\(=\left(\dfrac{1}{3}\cdot\sqrt[3]{3}+4\cdot\dfrac{1}{6}\cdot\sqrt[3]{3}-2\sqrt[3]{\dfrac{1}{2}}\right)\left(2\sqrt[3]{9}+2\sqrt[3]{12}+4\sqrt[3]{2}\right)\)

\(=\left(\sqrt[3]{3}-2\sqrt[3]{\dfrac{1}{2}}\right)\left(6\sqrt[3]{3}+2\sqrt[3]{12}+4\sqrt[3]{2}\right)\)

\(=6\cdot3+2\sqrt[3]{36}+4\sqrt[3]{6}-12\sqrt[3]{\dfrac{3}{2}}-4\sqrt[3]{6}-8\)

\(=10+12\sqrt[3]{\dfrac{1}{6}}-6\sqrt[3]{12}\)

\(\sqrt[3]{15\sqrt{3}-26}=\sqrt[3]{-\left(26-15\sqrt{3}\right)}\)

\(=-\sqrt[3]{8-3\cdot2^2\cdot\sqrt{3}+3\cdot2\cdot3-3\sqrt{3}}\)

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

giúp mình với mình đang cần gấp

giúp mình với

\(\left(\sqrt[3]{x}+1\right)^3-\left(\sqrt[3]{x}-1\right)^3-6\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x}+1\right)\\ =x+3\sqrt[3]{x^2}+3\sqrt[3]{x}+1-\left(x-3\sqrt[3]{x^2}+3\sqrt[3]{x}-1\right)-6\left(\sqrt[3]{x^2}-1\right)\\ =x+3\sqrt[3]{x^2}+3\sqrt[3]{x}+1-x+3\sqrt[3]{x^2}-3\sqrt[3]{x}+1-6\sqrt[3]{x^2}+6\\ =8\)

Lời giải:
Gọi biểu thức là $A$

\(A=(x+3\sqrt[3]{x^2}+3\sqrt[3]{x}+1)-(x-3\sqrt[3]{x^2}+3\sqrt[3]{x}-1)-6(\sqrt[3]{x^2}-1)\)

\(6\sqrt[3]{x^2}+2-6(\sqrt[3]{x^2}-1)=8\) là giá trị không phụ thuộc vào biến.