a) x^2 + 4y^2 + 6x - 12y + 18 = 0

<=>x²+6x+9+4y²-12y+9=0

<=>(x+3)²+(2y-3)²=0

<=>x+3=0 và 2y-3=0

<=>x=-3 và y=3/2

b) 5x^2 +9y^2 - 12xy - 6x +9 = 0

<=>x²-6x+9+4x²-12xy+9y²=0

<=>(x-3)²+(2x-3y)²=0

<=>x-3=0 và 2x-3y=0

<=>x=3 và 2.3-3y=0

<=>x=3 và y=2

                                                                   Bài giải

a, \(\frac{2}{7}x+\frac{1}{2}=-\frac{3}{4}\)

\(\frac{2}{7}x=-\frac{3}{4}-\frac{1}{2}\)

\(\frac{2}{7}x=-\frac{5}{4}\)

\(x=-\frac{5}{4}\text{ : }\frac{2}{7}\)

\(x=-\frac{35}{8}\)

b, \(\left(6x+\frac{2}{5}\right)=-\frac{8}{125}\)

\(6x=-\frac{8}{125}-\frac{2}{5}\)

\(6x=-\frac{58}{125}\)

\(x=-\frac{58}{125}\text{ : }6\)

\(x=\frac{-29}{375}\)

c, \(\left|x-\frac{2}{3}\right|\cdot\left(18-6x^2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\left|x-\frac{2}{3}\right|=0\\18-6x^2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=0\\6x^2=18\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x^2=3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\sqrt{3}\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{\frac{2}{3}\text{ ; }\sqrt{3}\right\}\)

a) ( x-34) x 15=0
   ( x-34)       =0 : 15
   ( x-34)       =0 
     x             =0 + 34
     x             =34
b) 18 x ( x-16)=18
          ( x- 16)=18 : 18
          ( x-16) =1
            x       = 16 + 1 
            x       = 17
c) ( x-4).( x-3)=0
   x . ( 4-3)    =0
   x . 0          =0 ( n* . o = 0 )
   => x = n*

e) 6.x + 4.x=2020
   10.x      =2020
       x      =2020:10
      x      =202

a)34

b)17

c)4

d)17

e)202

f)0

\(6x^2+\left(2y-1\right)x+10y^2-28y+18=0\)

\(\Delta=\left(2y-1\right)^2-24\left(10y^2-28y+18\right)\ge0\)

\(\Leftrightarrow-236y^2+668y-431\ge0\)

\(\Rightarrow\dfrac{167-2\sqrt{615}}{118}\le y\le\dfrac{167+2\sqrt{615}}{118}\)

\(\Rightarrow y=1\)

Thế vào pt đầu ...

a) ta có: \(-6x=18\)

\(\Rightarrow x=\frac{18}{-6}\)

\(\Rightarrow x=-3\)

b) ta có: \(2x-\left(-3\right)=7\)

\(\Rightarrow2x+3=7\)

\(\Rightarrow2x=7-3\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=2\)

c) ta có: \(\left(x-5\right)\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)

Vậy....

hok tốt!!

\(a,-6x=18\)

<=>\(x=18:\left(-6\right)\)

<=>\(x=-3\)

\(b,2x-\left(-3\right)=7\)

<=>\(2x+3=7\)

<=>\(2x=4\)

<=>\(x=2\)

\(c,\left(x-5\right)\left(x+6\right)=0\)

<=>\(\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)

A) -6x=18

        x=18:(-6)

        x=-3

Vậy x=-3

B) 2x -(-3)=7

     2x+3=7

     2x=7-3

     2x=4

       x=4:2

       x=2

vậy x=2

C) (x-5)(x+6)=0

* x-5=0            * x+6=0

      x=0+5              x=0-6

      x=5                  x=-6

Vậy x=5 hoặc x=-6

Bài 1 : 

\(a)\)\(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+3\right)\left(x-3\right)=15\)

\(\Leftrightarrow\)\(x^3-1-x\left(x^2-3^2\right)=15\)

\(\Leftrightarrow\)\(x^3-1-x^3+9x=15\)

\(\Leftrightarrow\)\(9x=16\)

\(\Leftrightarrow\)\(x=\frac{16}{9}\)

Vậy \(x=\frac{16}{9}\)

Chúc bạn học tốt ~ 

6x(3x + 5) - 2x(3x - 2) + (17 - x)(x - 1) + x(x - 18) = 0

=> (18x² - 6x² - x² + x²) + (30x + 4x - 16x - 18x) - 17 = 0

=> 12x² - 17 = 0

=> 12x² = 17

=> x² = 17/12

=> \(\orbr{\begin{cases}x=\sqrt{\frac{17}{12}}\\x=-\sqrt{\frac{17}{12}}\end{cases}}\)

\(6x\left(3x+5\right)-2x\left(3x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)=0\)

\(\Leftrightarrow9x^2+30x-6x^2+4x+17x-17-x^2+x+x^2-18x=0\)

\(\Leftrightarrow3x^2+34x-17=0\) ( vô nghiệm ) 

6x( 3x + 5 ) - 2x( 3x - 2 ) + ( 17 - x )( x - 1 ) + x( x - 18 ) = 0

<=> 18x² + 30x - 6x² + 4x - x² + 18x - 17 + x² - 18x = 0

<=> 12x² + 34x - 17 = 0

\(\Delta'=b'^2-ac=\left(\frac{b}{2}\right)^2-ac=\left(\frac{34}{2}\right)^2-12\cdot\left(-17\right)=289+204=493\)( không muốn xài Delta nữa đâu nhưng ... :)) )

\(\Delta'>0\)nên phương trình đã cho có hai nghiệm phân biệt

\(\hept{\begin{cases}x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-17+\sqrt{493}}{12}\\x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{-17-\sqrt{493}}{12}\end{cases}}\)

Vậy ...

       \(x^2+2y^2+2xy-6x+18=0\)

\(\Rightarrow\left(x+y\right)^2-6\left(x+y\right)+9+y^2+6y+9=0\)

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

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

\(\Rightarrow\hept{\begin{cases}x+y-3=0\\y+3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=6\\y=-3\end{cases}}\)

Cách khác :

\(x^2+2y^2+2xy-6x+18=0\)

\(2\left(x^2+2y^2+2xy-6x+18\right)=0\)

\(2x^2+4y^2+4xy-12x+36=0\)

\(\left[\left(2y\right)^2+2\cdot2y\cdot x+x^2\right]+\left(x^2-2\cdot x\cdot6+6^2\right)=0\)

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

\(\Rightarrow\hept{\begin{cases}2y+x=0\\x-6=0\end{cases}\Rightarrow\hept{\begin{cases}2y=-6\\x=6\end{cases}\Rightarrow}\hept{\begin{cases}y=-3\\x=6\end{cases}}}\)

P.s: Pham Van Hung đây là cách khác :)