\(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow\left(2x^3+4x^2\right)+\left(3x^2+6x\right)+\left(x+2\right)=0\)

\(\Leftrightarrow2x^2\left(x+2\right)+3x\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x^2+3x+1\right)=0\)

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

\(\Leftrightarrow\left(x+2\right)\left(x+1\right)\left(2x+1\right)=0\)

.......................................................................................

\(x^3-8x^2-8x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-8x\left(x+1\right)=0\)

......................................................................................

cảm ơn nha 

a:=>6x^2-8x+4x-6x^2<-4

=>-4x<-4

=>x>1

b: =>6x+8x^2-8x^2-24x>5

=>-18x>5

=>x<-5/18

a)\(6x^2-8x+2x\left(2-3x\right)< -4\)

\(\Leftrightarrow6x^2-8x+4x-6x^2< -4\)

\(\Leftrightarrow-4x< -4\)

\(\Leftrightarrow-4x.\dfrac{-1}{4}>-4\cdot\dfrac{-1}{4}\)

\(\Leftrightarrow x>1\)

Vậy bất phương trình có nghiệm là \(S=\left\{xIx>1\right\}\)

b)\(2\left(3x+4x^2\right)-8x\left(x+3\right)>5\)

\(\Leftrightarrow6x+8x^2-8x^2-24x>5\)

\(\Leftrightarrow-18x>5\)

\(\Leftrightarrow-18x\cdot\dfrac{-1}{18}< 5\cdot\dfrac{-1}{18}\)

\(\Leftrightarrow x< -\dfrac{5}{18}\)

Vậy bất phương trình có nghiệm là \(S=\left\{xIx< -\dfrac{5}{18}\right\}\)

cái này là tập nghiệm chứ bạn

c:

ĐKXĐ: 6-5x>=0

=>5x<=6

=>x<=1,2

\(2\sqrt[3]{3x-2}-3\cdot\sqrt{6-5x}+16=0\)

=>\(2\cdot\sqrt[3]{3x-2}+4+12-3\cdot\sqrt{6-5x}=0\)

=>\(2\cdot\left(\sqrt[3]{3x-2}+2\right)+3\left(4-\sqrt{6-5x}\right)=0\)

=>\(2\cdot\frac{3x-2+8}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{16-6+5x}{4+\sqrt{6-5x}}=0\)

=>\(2\cdot\frac{3x+6}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5x+10}{4+\sqrt{6-5x}}=0\)

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

=>x+2=0

=>x=-2(nhận)

d: ĐKXĐ: x>=1

\(\sqrt[3]{x+6}-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2-2\cdot\sqrt{x-1}+2=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)=\left(2-x\right)\left(2+x\right)\)

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

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

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

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

=>x-2=0

=>x=2(nhận)

Ta có : 

\(3x^3-8x^2-2x+4=\left(3x-2\right)\left(x^2-2x-2\right)\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x^2-2x-2=0\end{cases}}\)

Th1 : \(3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)

Th2: \(x^2-2x-2=0\)

\(\Leftrightarrow x^2-2x+1=3\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}}\)

Vậy phương trình có 3 nghiệm : \(x=1\), \(x=1\pm\sqrt{3}\)

\(3x^3-8x^2-2x+4=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x^2-2x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=1\pm\sqrt{3}\end{cases}}\)

Vậy tập nghiệm của phương trình \(S=\left\{\frac{2}{3};1\pm\sqrt{3}\right\}\)

a: \(x^4=5x^2+2x-3\)

=>\(x^4-5x^2-2x+3=0\)

=>\(x^4+x^3-x^2-x^3-x^2+x-3x^2-3x+3=0\)

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

TH1: \(x^2+x-1=0\)

=>\(x^2+x+\frac14=\frac54\)

=>\(\left(x+\frac12\right)^2=\frac54\)

=>\(x+\frac12=\pm\frac{\sqrt5}{2}\)

=>\(x=-\frac12\pm\frac{\sqrt5}{2}\)

TH2: \(x^2-x-3=0\)

=>\(x^2-x+\frac14-\frac{13}{4}=0\)

=>\(\left(x-\frac12\right)^2=\frac{13}{4}\)

=>\(x-\frac12=\pm\frac{\sqrt{13}}{2}\)

=>\(x=\frac12\pm\frac{\sqrt{13}}{2}\)

c: \(3x^3+3x^2+3x=-1\)

=>\(x^3+3x^2+3x+1=-2x^3\)

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

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

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

=>\(x=\frac{-1}{1-\sqrt[3]{-2}}\)

d: \(8x^3-12x^2+6x-5=0\)

=>\(8x^3-12x^2+6x-1-4=0\)

=>\(\left(2x-1\right)^3=4\)

=>\(2x-1=\sqrt[3]{4}\)

=>\(2x=1+\sqrt[3]{4}\)

=>\(x=\frac12+\frac12\cdot\sqrt[3]{4}\)

b)\(3x^3+6x^2-75x-150=0\Leftrightarrow3\left(x^3+2x^2-25x-50\right)=0\Leftrightarrow x^3+2x^2-25x-50=0\)

<=>\(x^2\left(x+2\right)-25\left(x+2\right)=0\Leftrightarrow\left(x^2-25\right)\left(x+2\right)=0\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+2\right)=0\)

<=>x-5=0 hoặc x+5=0 hoặc x+2=0<=>x=5 hoặc x=-5 hoặc x=-2

c)\(2x^5-3x^4+6x^3-8x^2+3=0\Leftrightarrow2x^5+x^4-4x^4-2x^3+8x^3+4x^2-12x^2+3=0\)

<=>\(x^4\left(2x+1\right)-2x^3\left(2x+1\right)+4x^2\left(2x+1\right)-3\left(4x^2-1\right)=0\)

<=>\(x^4\left(2x+1\right)-2x^3\left(2x+1\right)+4x^2\left(2x+1\right)-3\left(2x-1\right)\left(2x+1\right)=0\)

<=>\(\left(2x+1\right)\left(x^4-2x^3+4x^2-6x+3\right)=0\)

<=>\(\left(2x+1\right)\left(x^4-2x^3+x^2+3x^2-6x+3\right)=0\)

<=>\(\left(2x+1\right)\left[x^2\left(x^2-2x+1\right)+3\left(x^2-2x+1\right)\right]=0\)

<=>\(\left(2x+1\right)\left(x^2+3\right)\left(x^2-2x+1\right)=0\Leftrightarrow\left(2x+1\right)\left(x^2+3\right)\left(x-1\right)^2=0\)

Vì \(x^2\ge0\Rightarrow x^2+3\ge3>0\Rightarrow\orbr{\begin{cases}2x+1=0\\\left(x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

a) 2x³ - x² - 8x + 4 = 0

x².(2x - 1) - 4.(2x - 1) = 0

(x² - 4)(2x - 1) = 0

\(\Rightarrow\orbr{\begin{cases}x^2-4=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=\frac{1}{2}\end{cases}}\)

Với x² = 4

=> x = 2 hoặc x = -2

=> x = {-2 ; 2 ; \(\frac{1}{2}\))