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

a) x( x + 3 ) - 2x - 6 =0

=> x^2 + 3x - 2x -6 = 0

=> x^2 + x - 6 = 0

=> ( x^2 -x ) + ( 6x - 6 ) = 0

=> x( x - 1 ) + 6( x - 1 ) = 0

=> ( x - 1 )( x + 6 ) = 0

=> x = 1 hoặc x= -6

b) 9x^2 - 6x - 3 = 0

=> (9x^2 + 3x ) - ( 9x + 3 ) = 0

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

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

=> x = -1/3 hoặc x = 1 

Vũ ơi! Dòng thứ 3 xuống dòng thứ 4 câu a. Em phân tích bị sai rồi. Em có thể làm theo cách khác mà  không cần phân tích ra không ? Sử dụng -2x - 6  = - 2 ( x + 3 )

Câu b. Đúng rồi. 

a) x(x + 3) - 2x - 6 = 0

<=> x^2 + 3x - 2x - 6 = 0

<=> x^2 + x - 6 = 0

<=> (x - 2)(x + 3) = 0

<=> x - 2 = 0 hoặc x + 3 = 0

=> x = 2 hoặc x = -3

b) 9x^2 - 6x - 3 = 0

<=> 3(3x^2 - 2x - 1) = 0

<=> 3(3x^2 + x - 3x - 1) = 0

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

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

<=> 3x + 1 = 0 hoặc x - 1 = 0

=> x = -1/3 hoặc x = 1

d) \(4x^2-9-x\left(2x-3\right)=0\)

\(\Leftrightarrow4x^2-9-2x^2+3x=0\)

\(\Leftrightarrow2x^2+3x-9=0\)

\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)

Vậy pt có 2 nghiệm phân biệt

\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)

e) \(x^3+5x^2+9x=-45\)

\(\Leftrightarrow x^3+5x^2+9x+45=0\)

\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)

\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)

f) \(x^3-6x^2-x+30=0\)

\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)

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

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

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

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

\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)

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

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

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

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

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

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 ...

9x²-6x-3=0

=>9x²-9x+3x-3=0

=>(x-1)(9x-3)=0

=>x-1=0 hoặc 9x+3 = 0

=> x=1 hoặc x=-1/3

b. x³+9x²+27x+19=0

   x³+x²+8x²+8x+19x+19=0

(x+1)(x²+8x+19)=0

x+1=0 => x=-1 

x²+8x+19= x²+8x+16+3=(x+4)²+3 lớn hơn hoặc bằng 3., lớn hơn 0 với moị x

a, \(\Rightarrow3\left(3x^2-2x-1\right)=0\)

\(\Rightarrow3x^2-2x-1=0\)

\(\Rightarrow x\left(3x-2\right)=1\)

\(\Rightarrow\orbr{\begin{cases}x=1\\3x-2=1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\3x-2=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

b,\(\Rightarrow x^3+3x^2+6x^2+9x+18x+19=0\)

\(\Rightarrow x^2\left(x+3\right)+3x\left(x+3\right)+18\left(x+3\right)-2=0\)

\(\Rightarrow\left(x+3\right)\left(x^2+3x+18\right)=2\)

Mk k co thoi gian. buoc tiep theo tu lam not nhe

<=>\(9x^2-6x+1-4=0\)

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

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

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

<=>3x-1=2

<=>3x=3

<=>x=1

Vậy x=1

\(\frac{x-1}{x^2-9x+20}+\frac{2x-2}{x^2-6x+8}+\frac{3x-3}{x^2-x-2}+\frac{4x-4}{x^2+6x+5}=0\)

\(\Leftrightarrow\frac{x-1}{\left(x-5\right)\left(x-4\right)}+\frac{2\left(x-1\right)}{\left(x-4\right)\left(x-2\right)}+\frac{3\left(x-1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{10}{x^2-25}\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)  

PS: Điều kiện xác đinh bạn tự làm nhé 

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: \(5x\left(x-3\right)-x+3=0\)

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

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

\(a,9x^2-6x-3=0\)

\(\Leftrightarrow9x^2-6x+1-4=0\)

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

\(\Rightarrow3x-1=\pm2\)

\(\hept{\begin{cases}3x-1=2\Rightarrow x=1\\3x-1=-2\Rightarrow x=\frac{-1}{3}\end{cases}}\)

Vậy \(x=1\) hoặc \(x=\frac{-1}{3}\)

\(b,x^3+9x^2+27x+19=0\)

\(\Leftrightarrow x^3+9x^2+27x+27-8=0\)

\(\Leftrightarrow\left(x+3\right)^3=8\)

\(\Rightarrow x+3=2\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

\(c,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)

\(\Leftrightarrow x^3-25x-x^3-8=3\)

\(\Leftrightarrow-25x=11\)

\(\Leftrightarrow x=\frac{-11}{25}\)

Vậy \(x=\frac{-11}{25}\)

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

<=> \(\left(3x\right)^2-2.3x.1+1-4=0\)

<=> \(\left(3x-1\right)^2-2^2=0\)

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

<=> \(\hept{\begin{cases}3x-3=0\\3x+1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)

\(x^3+9x^2+27x+19\) \(=0\)

<=>\(x^3+x^2+8x^2+8x+19x+19=0\)

<=> \(x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)

<=> \(\left(x^2+8x+19\right)\left(x+1\right)=0\)

mà \(x^2+8x+19>0\)

=> \(x+1=0\)

<=> \(x=-1\)

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

<=> \(x\left(x^2-25\right)-\left(x+2\right)\left(x-2\right)^2=3\)

<=> \(x^3-25x-\left(x^2-4\right)\left(x-2\right)=3\)

<=>  \(x^3-25x-\left(x^3-2x^2-4x+8\right)=3\)

<=> \(x^3-25x-x^3+2x^2+4x-8=3\)

<=> \(2x^2-21x-8=3\)

<=> \(2x^2-21x-11=0\)

<=> \(2x^2-22x+x-11=0\)

<=> \(2x\left(x-11\right)+\left(x-11\right)=0\)

<=> \(\left(2x+1\right)\left(x-11\right)=0\)

<=> \(\hept{\begin{cases}2x+1=0\\x-11=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=11\end{cases}}\)