\(x^3-9x+7x^2-63=0\)

\(\Rightarrow\left(x^3+7x^2\right)-9x-63=0\)

\(\Rightarrow x^2\left(x+7\right)-9\left(x+7\right)=0\)

\(\Rightarrow\left(x^2-9\right)\left(x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-9=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=9\\x=-7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\pm3\\x=-7\end{cases}}}\)

Vậy ...

x3−9x+7x2−63=0x3−9x+7x2−63=0

⇒(x3+7x2)−9x−63=0⇒(x3+7x2)−9x−63=0

⇒x2(x+7)−9(x+7)=0⇒x2(x+7)−9(x+7)=0

⇒(x2−9)(x+7)=0⇒(x2−9)(x+7)=0

⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7⇒{x2−9=0x+7=0⇒{x2=9x=−7⇒{x=±3x=−7

Vậy ...

Mình trình bày trong hình ^^ Bn tham khảo nhé

d: Ta có: \(9x^2+6x-8=0\)

\(\Leftrightarrow9x^2+12x-6x-8=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)

e: Ta có: \(x\left(x-2\right)+x-2=0\)

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

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

f: Ta có: \(5x\left(x-3\right)-x+3=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

a) x³-9x²+27x-27=0

<=>(x-3)³=0

<=>x-3=0

<=>x=3

b) x³-25x=0

<=>x.(x²-25)=0

<=>x.(x-5)(x+5)=0

<=>x=0 hoặc x-5=0 hoặc x+5=0

<=>x=0 hoặc x=5 hoặc x=-5

c)9x²-1=0

<=>(3x-1)(3x+1)=0

<=>3x-1=0 hoặc 3x+1=0

<=>x=1/3 hoặc x=-1/3

a, x^3 - 9x^2 + 27x - 27 = 0 

=> ( x - 3)^3 = 0 

=> x - 3 = 0 

=> x = 3 

b, x^3 - 25x = 0 

=> x(x^2 - 25) = 0 

=> x(x-5)(x + 5) = 0 

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

=> x= 0 hoặc x =5 hoặc x = -5 

c, 9x^2 -  1 = 0 

 => (3x)^2 - 1^2 = 0 

=> ( 3x- 1)(3x+ 1) = 0 

=> 3x - 1 = 0 hoặc 3x + 1 = 0 

=> x = 1/3 hoặc x = -1/3  

a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x\right)+17=0\Leftrightarrow5\left(x^2-2.\frac{2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+17=0\)

\(\Leftrightarrow5\left(x-\frac{2}{5}\right)^2-\frac{4}{5}+17=0\Leftrightarrow5\left(x-\frac{2}{5}\right)^2+81\ge81>0\)

Vậy pt vô nghiệm 

b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\Leftrightarrow x.2x=0\Leftrightarrow x=0\)

c, \(2x^2-9x+7=0\Leftrightarrow2x^2-7x-2x+7=0\)

\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\Leftrightarrow\left(x-1\right)\left(2x-7\right)=0\Leftrightarrow x=1;x=\frac{7}{2}\)

Trả lời:

a, \(5x\left(x-1\right)+\left(x+17\right)=0\)

\(\Leftrightarrow5x^2-5x+x+17=0\)

\(\Leftrightarrow5x^2-4x+17=0\)

\(\Leftrightarrow5\left(x^2-\frac{4}{5}x+\frac{17}{5}\right)=0\)

\(\Leftrightarrow x^2-\frac{4}{5}x+\frac{17}{5}=0\)

\(\Leftrightarrow x^2-2.x.\frac{2}{5}+\frac{4}{25}+\frac{81}{25}=0\)

\(\Leftrightarrow\left(x-\frac{2}{5}\right)^2+\frac{81}{25}=0\)

Vì \(\left(x-\frac{2}{5}\right)^2+\frac{81}{25}\ge\frac{81}{25}>0\forall x\)

nên pt vô nghiệm 

b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)

\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)

\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\)

\(\Leftrightarrow3x.\left(-9\right).2x=0\)

\(\Leftrightarrow-54x^2=0\)

\(\Leftrightarrow x^2=0\)

\(\Leftrightarrow x=0\)

Vậy x = 0 là nghiệm của pt.

c, \(7-9x+2x^2=0\)

\(\Leftrightarrow2x^2-7x-2x+7=0\)

\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}}\)

Vậy x = 7/2; x = 1 là nghiệm của pt.

d, trùng ý c

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

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

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

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)

3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)

4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)

\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)

5, em xem lại đề nhé

à lag tý @@

5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)

a)\(x^3+4x^2+4x=0\)

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

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

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

b)\(\left(x+3\right)^2-4=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3-2=0\\x+3+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}}\)

c)\(x^4-9x^2=0\)

\(\Leftrightarrow x^2\left(x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

d)\(x^2-6x+9=81\)

\(\Leftrightarrow\left(x-3\right)^2=81\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)

e)\(x^3+6x^2+9x-4x=0\)

\(\Leftrightarrow x^3+6x^2+5x=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0;x=-5\\x=-1\end{cases}}}\)

#H

( x - 2 ) ³ - ( x + 1 ) ³ + 9x ( x + 1 ) - 9 = 0
=> \(x^3-6x^2+12x-8-\left(x^3+3x^2+3x+1\right)+9x^2+9x-9=0\)
=> \(x^3-6x^2+12x-8-x^3-3x^2-3x-1+9x^2+9x-9=0\)
=> \(18x-18=0\)
=>         \(18x=0+18\)
=>         \(18x=18\)
=>             \(x=1\)

a/ => x² + 3x - 10  = 0

=> x² - 2x + 5x - 10 = 0

=> x(x - 2) + 5(x - 2) = 0

=> (x - 2)(x + 5) = 0 

=> x - 2 = 0 => x = 2

hoặc x + 5 = 0 => x = -5

Vậy x = 2; x = -5

b/ => 3(x³ + 3x² + 3x + 1) = 0 

=> x³ + 3x² + 3x + 1 = 0 

=> (x + 1)³ = 0 

=> x + 1 = 0 => x = -1

Vậy x = -1

\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)

\(\Leftrightarrow\left|x\left(x^2+1\right)\right|-9\left|x^2+1\right|=0\)

\(\Leftrightarrow\left(\left|x\right|-9\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\left|x\right|=9\left(x^2+1\ge1>0\right)\Leftrightarrow x=\pm9\)

Vậy ... 

\(\left|x^3+x\right|-\left|9x^2+9\right|=0\)

\(TH1:\left\{{}\begin{matrix}\left|x^3+x\right|=0\\\left|9x^2+9\right|=0\end{matrix}\right.\)

\(\text{Vì }9x^2\ge0\)

\(\Rightarrow9x^2+9\ge9\)

\(TH2:\left|x^3+x\right|=\left|9x^2+9\right|\)

\(\Rightarrow\left[{}\begin{matrix}x^3+x=9x^2-9\\x^3+x=9x^2+9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^3+x+9x^2+9=0\\x^3+x-9x^2-9=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x.\left(x^2+1\right)+9.\left(x^2+1\right)=0\\x.\left(x^2+1\right)-9.\left(x^2+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=9\end{matrix}\right.\)

=>|x^3+x|=|9x^2+9|

=>x^3+x=9x^2+9 hoặc x^3+x=-9x^2-9

=>x^3-9x^2+x-9=0 hoặc x^3+9x^2+x+9=0

=>x+9=0 hoặc (x-9)(x^2+1)=0

=>x=9 hoặc x=-9