a ) 

\(5x\left(x-3\right)+7\left(x-3\right)=0\)

\(\Rightarrow\left(5x+7\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+7=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-7\\x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{7}{5}\\x=3\end{cases}}}\)

Vậy ...

b ) 

\(x^{2017}=x^{2018}\)

\(\Rightarrow x^{2017}-x^{2018}=0\)

\(\Rightarrow x^{2017}\left(1-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^{2017}=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy ...

c ) 

\(2x^2=x\)

\(\Rightarrow2x^2:x=1\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy ...

e ) 

\(x^5=x^4\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)( làm tương tự như phần b )  

~ 

d ) 

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

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

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

Vậy ...

sao dễ zữ lun vậy

nhok sư tử: Dễ thì làm đi

\(c.\left(x+7\right).\left(x+8\right)=0\)

\(=>x+7=0\)hoặc \(x+8=0\)

\(=>x=0-7=-7\)hoặc \(x=0-8=-8\)

Vậy x = -7 hoặc-8

Bài 1 : 

a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)

TH1 : \(x-3=2\Leftrightarrow x=5\)

TH2 : \(x-3=-2\Leftrightarrow x=1\)

b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)

TH1 : \(x-6=0\Leftrightarrow x=6\)

TH2 : \(x+4=0\Leftrightarrow x=-4\)

c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)

\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)

d, tương tự 

Bài 2 :

 \(x^2+2xy+y^2-6x-6y-5=\left(x+y\right)^2-6\left(x+y\right)-5\)

Thay x + y = -9 ta có : 

\(\left(-9\right)^2-6\left(-9\right)-5=130\)

Bài 1.

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² - 2² = 0

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

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

<=> \(\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) x² - 2x = 24

<=> x² - 2x - 24 = 0

<=> x² + 4x - 6x - 24 = 0

<=> x( x + 4 ) - 6( x + 4 ) = 0

<=> ( x + 4 )( x - 6 ) = 0

<=> \(\orbr{\begin{cases}x+4=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)

c) ( 2x - 1 )² + ( x + 3 )² - 5( x + 2 )( x - 2 ) = 0

<=> 4x² - 4x + 1 + x² + 6x + 9 - 5( x² - 4 ) = 0

<=> 5x² + 2x + 10 - 5x² + 20 = 0

<=> 2x + 30 = 0

<=> 2x = -30

<=> x = -15

Bài 2.

D = x² + 2xy + y² - 6x - 6y - 5

= [ ( x² + 2xy + y² ) - 2x - 2y + 1 ] - 4x - 4y - 6

= [ ( x + y )² - 2( x + y ) + 1 ] - 4( x + y ) - 6

= ( x + y - 1 )² - 4( x + y ) - 6

Với x + y = -9

D = ( -9 - 1 )² - 4.(-9) - 6

    = 100 + 36 - 6

    = 130

Bài 3.

4x² + y² - 4x + 10y + 26 = 0

<=> ( 4x² - 4x + 1 ) + ( y² + 10y + 25 ) = 0

<=> ( 2x - 1 )² + ( y + 5 )² = 0

<=> \(\hept{\begin{cases}2x-1=0\\y+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-5\end{cases}}\)

a) 4x(x+1)=8(x+1)

<=>4x(x+1)-8(x+1)=0

<=>(4x-8)(x+1)=0

<=>\(\left[\begin{array}{} 4x-8=0\\ x+1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=2\\ x=-1 \end{array} \right.\)

Vậy...

b)x(x-1)-2(1-x)=0

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

<=>\(\left[\begin{array}{} x+2=0\\ x-1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-2\\ x=1 \end{array} \right.\)

Vậy...

c)5x(x-2)-(2-x)=0

<=>(5x+1)(x-2)=0

<=>\(\left[\begin{array}{} 5x+1=0\\ x-2 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-1/5\\ x=2 \end{array} \right.\)

d)5x(x-200)-x+200=0

<=>(5x-1)(x-200)=0

<=>\(\left[\begin{array}{} 5x-1=0\\ x-200=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=1/5\\ x=200 \end{array} \right.\)

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

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

\(\Leftrightarrow \left[\begin{array}{} x=0\\ x^2+4=0 (loại vì x^2+4>=0 với mọi x) \end{array} \right.\)

Vậy x=0

f)\((x+1)=(x+1)^2\)

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

\(\Leftrightarrow (x+1)(1-x-1)=0\)

\(\Leftrightarrow (x+1)(-x)=0\)

\(\Leftrightarrow \left[\begin{array}{} x=-1\\ x=0 \end{array} \right.\)

Vậy....

a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

d) 5x(12x + 7) – 3x(20x – 5) = - 100

⇒60x²+35x-60x²+15=-100

⇒35x+15=-100

⇒35x=-100-15

⇒35x=-115

⇒x=\(\dfrac{-115}{35}\)

⇒x=\(\dfrac{-23}{7}\)

ngu thế à bạn

a) \(\Rightarrow x\left(x+3\right)=0\)

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

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

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

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

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

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow80x=480\Rightarrow x=6\)

b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow4x=0\Rightarrow x=0\)

c) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

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

b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)