a)\(\sqrt[3]{x-5}+\sqrt[3]{x+2}=3\left(ĐKXĐ:x\in R\right)\)

\(\Leftrightarrow\left(\sqrt[3]{x-5}-1\right)+\left(\sqrt[3]{x+2}-2\right)=0\)

\(\Leftrightarrow\frac{x-5-1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{x+2-8}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\)

\(\Leftrightarrow\frac{x-6}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{x-6}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\)

\(\Leftrightarrow\left(x-6\right)\left[\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}\right]=0\)

a') (tiếp) 

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\left(TMĐKXĐ\right)\\\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\end{cases}}\)

Xét phương trình:

\(\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\left(1\right)\)

Ta có:

\(\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1=\left(\sqrt[3]{x-5}+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\in R\)

\(\Rightarrow\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}>0\forall x\in R\)

a'') (tiếp)

\(\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{\left(x+2\right)}+4=\left(\sqrt[3]{x+2}+1\right)^2+3>0\forall x\in R\)\(\Rightarrow\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}>0\forall x\in R\)

Suy ra \(\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}>0\forall x\in R\)

Do đó phương trình (1) vô nghiệm.

Vậy phương trình có nghiệm duy nhất: \(x=6\).

b) \(\sqrt[3]{x}+\sqrt{x+1}=5\left(ĐKXĐ:x\ge-1\right)\)

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

\(\Leftrightarrow\frac{x-8}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\frac{x+1-9}{\sqrt{x+1}+3}=0\Leftrightarrow\frac{x-8}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\frac{x-8}{\sqrt{x+1}+3}=0\)

\(\Leftrightarrow\left(x-8\right)\left(\frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\frac{1}{\sqrt{x+1}+3}\right)=0\)

b') (tiếp)

\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\\frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\frac{1}{\sqrt{x+1}+3}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=8\left(TMĐKXĐ\right)\\\frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\frac{1}{\sqrt{x+1}+3}=0\end{cases}}\)

Xét phương trình \(\frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\frac{1}{\sqrt{x+1}+3}=0\)(1)

Ta có:

\(\sqrt[3]{x^2}+2\sqrt[3]{x}+4=\left(\sqrt[3]{x}+1\right)^2+3>0\forall x\ge-1\)(\(x\ge-1\)theo ĐKXĐ)

b'') (tiếp)

\(\Rightarrow\frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}>0\forall x\ge-1\);

\(\sqrt{x+1}\ge0\forall x\ge-1\)

\(\Rightarrow\sqrt{x+1}+3\ge3>0\forall x\ge-1\)

\(\Rightarrow\frac{1}{\sqrt{x+1}+3}>0\forall x\ge-1\)

Suy ra \(\frac{1}{\sqrt[3]{x^2}+2\sqrt[3]{x}+4}+\frac{1}{\sqrt{x+1}+3}>0\forall x\ge-1\)

Do đó phương trình (1) vô nghiệm.

Vậy phương trình có nghiệm duy nhất : x = 8.

...

Dạ, kiểu bài này em sử dụng Công thức toán học không quá 5 lần ạ.

 b) \(\sqrt[3]{x}+\sqrt{x+1}=5\)  (2) 

đặt a=\(\sqrt[3]{x}\) ; b=\(\sqrt{x+1}\) được a+b=5

 ⇒ b²-a³=( \(\sqrt{x+1}\))²-(\(\sqrt[3]{x}\))³

             = x+1-x

            =1

a) \(\sqrt[3]{x-5}\)+\(\sqrt[3]{x+2}\)=3

đặt a= \(\sqrt[3]{x-5}\)  ; b=\(\sqrt[3]{x+2}\) được a+b=3⇒a=3-b⇒a³=(3-b)³⇒a³= 27-27b+9b²-b³

+) b³-a³= (x+2)-(x-5)=x+2-x+5=7  (1) 

thay a³=27-27b+9b²-b³ vào (1) ta có 

b³-a³=7

⇔b³-( 27-27b+9b²-b³) =7

⇔b³-27+27b-9b²+b³=7

⇔2b³+27b-9b²=7+27