a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² = 4

<=> \(\orbr{\begin{cases}\left(x-3\right)^2=2^2\\\left(x-3\right)^2=\left(-2\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

<=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

Vậy S = { 5 ; 1 }

b) x² - 9 = 0

<=> x² = 9

<=> \(\orbr{\begin{cases}x^2=3^2\\x^2=\left(-3\right)^2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy S = { 3 ; -3 }

c) x( x - 2x ) - x² - 8 = 0

<=> x² - 2x² - x² - 8 = 0

<=> -2x² - 8 = 0

<=> -2x² = 8

<=> x² = -4 ( vô lí )

<=> x = \(\varnothing\)

Vậy S = { \(\varnothing\)}

d) 2x( x - 1 ) - 2x² + x - 5 = 0

<=> 2x² - 2x - 2x² + x - 5 = 0

<=> -x - 5 = 0

<=> -x = 5

<=> x = -5

Vậy S = { -5 }

e) x( x - 3 ) - ( x + 1 )( x - 2 ) = 0 

<=> x² - 3x - ( x² - x - 2 ) = 0

<=> x² - 3x - x² + x + 2 = 0

<=> - 2x + 2 = 0

<=> -2x = -2

<=> x = 1

Vậy S = { 1 }

f) x( 3x - 1 ) - 3x² - 7x = 0

<=> 3x² - x - 3x² - 7x = 0

<=> -8x = 0

<=> x = 0

Vậy S = { 0 } 

d) 2x(x - 1) - 2x² + x - 5 = 0

=> 2x² - 2x - 2x² + x - 5 = 0

=> -x = 5

=> x = -5

e) x(x - 3) - (x + 1)(x - 2) = 0

=> x² - 3x - (x² - x - 2) = 0 

=> x² - 3x - x² + x + 2 = 0

=> -2x = - 2

=> x = 1

f) x(3x - 1) - 3x² - 7x = 0

=> 3x² - x - 3x² - 7x = 0

=> -8x = 0

=> x = 0

c) \(x^2-9=2\cdot\left(x+3\right)^2\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)

b) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

d) \(x^2-8x+3x-24=0\)

\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)

\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)

a) \(x^2-9=2\left(x+3\right)^2\)

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

\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)

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

\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)

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

b) \(x^2-8x+3x-24=0\)

\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

c) \(x^3-3x^2+3x-1=0\)

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

a) 15x²-3x=0

=>3x(5x-1)=0

=>2 TH

=>*3x=0                   *5x-1=0

=>x=0                        =>5x=1=>x=1/5

vậy x=0 hoặc x=1/5

b) (3x-2) (x+3)+ (x²-9)=0

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

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

=>(x+3).(4x-5)=0

=> 2 TH

*x+3=0=>x=0-3=>x=-3

*4x-5=0=>4x=5=>x=5/4

vậy x=-3 hoặc x=5/4

c) (x-1)³- (x+1) (2-3x)=-3

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

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

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

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

\(\Rightarrow x^3+4x=0\)

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

=> 2 TH

*x=0

*x^2+4=0

vì: x^2>0

do đó:x^2+4>0

=> x^2+4 ko có gt nào x t/m y/cầu đề bài

vậy x=0

a) x = 1; x = - 1 3                 b) x = 2.

c) x = 3; x = -2.                 d) x = -3; x = 0; x = 2.

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

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

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

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

\(x.\left(x-3\right)-3x+9=0\)

\(x.\left(x-3\right)-3.\left(x-3\right)=0\)

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

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

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

\(\Leftrightarrow x^2+x-6=0\)

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

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

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

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

Nhiều qua trời 

mình đang cần rất gấp mong các bạn giúp ,ngày mai mình phải nộp cho cô rồi .bạn nào làm nhanh mình k luôn nha

\(\left(x+4\right).\left(-3x+9\right)=\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\-3x+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0-4\\-3x=0-9\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=-4\\-3x=-9\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\left(-9\right):\left(-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=3\end{cases}}}\)

Vậy .....................

~ Hok tốt ~

#)Giải ;

a)\(\left(x+4\right)\left(-3x+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+4=0\\-3x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-4\\-3x=-9\end{cases}\Rightarrow}\orbr{\begin{cases}x=-4\\x=3\end{cases}}}\)

Vậy \(x\in\left\{-4;3\right\}\)

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

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

Vậy \(x\in\left\{-1;5\right\}\)

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

\(\Rightarrow\orbr{\begin{cases}\left|x\right|+2=0\\x^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}\left|x\right|=-2\\x^2=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=2;x=-2\\x=1\end{cases}}}\)

Vậy \(x\in\left\{2;-2;1\right\}\)

d)\(\left(x^2-3\right)\left(2x^2+10\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-3=0\\2x^2+10=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=3\\2x^2=-10\end{cases}\Rightarrow}\orbr{\begin{cases}x=\sqrt{3}\\x^2=-5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\sqrt{3}\\x=\sqrt{-5}\end{cases}}}\)

Vậy \(x\in\left\{\sqrt{3};\sqrt{-5}\right\}\)

Làm 1 câu thuii nha mik nhát quá!! nhưng các bài còn lại tương tự nha!!

a. \(\left(x+1\right)\left(3-x\right)=0\)

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

Vậy..

hok tốt!!

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

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

vậy x=-1 hoặc x=3

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

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

vậy x=2 hoặc x=1/2

câu c tương tự

\(\left(x^2+1\right)\left(81-x^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\81-x^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x^2=81\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x\in\varnothing\\x=\pm9\end{cases}}\)

vậy x=9 hoặc x=-9

Những cách dành cho trâu bò xD

\(a,\left(x+1\right)\left(3-x\right)=0\)

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

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

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

Ta có : \(\Delta=2^2-4.\left(-1\right).3=4+12=16\)

Vì \(\Delta>0\)nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{-2+\sqrt{16}}{-2}=\frac{2}{-2}=-1\)

\(x_2=\frac{-2-\sqrt{16}}{-2}=\frac{-6}{-2}=3\)

Vậy ...

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đù khó thế

tl j z mấy chế , k câu dc đâu :))