Bài 9: Căn bậc ba

a: =>2x+1=27

=>2x=26

=>x=13

b: =>\(\sqrt[3]{x+5}=x+5\)

=>x+5=(x+5)^3

=>(x+5)(x+4)(x+6)=0

=>x=-5;x=-4;x=-6

c: =>2-3x=-8

=>3x=10

=>x=10/3

d: =>\(\sqrt[3]{x-1}=x-1\)

=>(x-1)^3=(x-1)

=>x(x-1)(x-2)=0

=>x=0;x=1;x=2

6B.
a)\(\sqrt[3]{4-2x}\ge4\Leftrightarrow4-2x\ge64\)
\(\Leftrightarrow2x\le-60\Leftrightarrow x\le-30\)
Vậy...

b) \(\sqrt[3]{-x^3-3x^2+6x-10}< -x-1\)

\(\Leftrightarrow-x^3-3x^2+6x-10< -\left(x+1\right)^3\)

\(\Leftrightarrow-x^3-3x^2+6x-10< -x^3-3x^2-3x-1\)

\(\Leftrightarrow9x< 9\Leftrightarrow x< 1\)
Vậy...
7A.
a) \(\sqrt[3]{2x+1}=3\Leftrightarrow2x+1=27\Leftrightarrow x=13\)
Vậy...
b) \(\sqrt[3]{5+x}-x=5\)

\(\Leftrightarrow5+x=\left(5+x\right)^3\) \(\Leftrightarrow\left[{}\begin{matrix}5+x=0\\\left(5+x\right)^2=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-4\\x=-6\end{matrix}\right.\)

Vậy...

7B.

a) PT \(\Leftrightarrow2-3x=-8\Leftrightarrow x=\dfrac{10}{3}\)

b) PT \(\Leftrightarrow x-1=\left(x-1\right)^3\) 

\(\Leftrightarrow\left(x-1\right)\left[1-\left(x-1\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\\x=2\end{matrix}\right.\)

Vậy...

\(\sqrt[3]{2x+1}>-5\)

\(\Leftrightarrow2x+1>\left(-5\right)^3\)

\(\Leftrightarrow2x+1>-125\)

\(\Leftrightarrow2x>-125-1\)

\(\Leftrightarrow2x>-126\)

\(\Leftrightarrow x>-\dfrac{126}{2}\)

\(\Leftrightarrow x>-63\)

\(\Leftrightarrow3\sqrt{2}\cdot x>-6\)

=>\(x>-\dfrac{6}{3\sqrt{2}}=-\sqrt{2}\)

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

\(\left(1\right)\Leftrightarrow x+2x-3+3.\sqrt[3]{x\left(2x-3\right)}.\left(\sqrt[3]{x}+\sqrt[3]{2x-3}\right)=12x-12\)

\(\Rightarrow\sqrt[3]{12x\left(x-1\right)\left(2x-3\right)}=3x-3\)

\(\Leftrightarrow12x\left(x-1\right)\left(2x-3\right)=[3\left(x-1\right)]^3\)

\(\Leftrightarrow12x\left(2x^2-5x+3\right)=27\left(x^3-3x^2+3x-1\right)\)

\(\Leftrightarrow24x^3-60x^2+36x=27x^3-81x^2+81x-27\)

\(\Leftrightarrow3x^3-21x^2+45x-27=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x-3\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\). Thử lại ta thấy cả x=1;x=3 đều t/m bài toán

Vậy, pt có tập nghiệm là S={1;3}

Để pt có hai nghiệm phân biệt trái dấu thì -m+5<0

=>-m<-5

=>m>5