x² + y² + 10x + 6y + 34 = 0

=> (x² + 10x + 25) + (y² + 6y + 9) = 0

=> (x + 5)² + (y + 3)² = 0

=> \(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)

Vậy x = - 5 ; y = -3

b) 25x² + 4y² + 10x + 4y + 2 = 0

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

=> (5x + 1)² + (2y + 1)² = 0

=> \(\hept{\begin{cases}5x+1=0\\2y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-0,2\\y=-0,5\end{cases}}\)

Vậy x = -0,2 ; y = -0,5

a) 

\(x^2+10x+25+y^2+6y+9=0\)    

\(\left(x+5\right)^2+\left(y+3\right)^2=0\)  ( 1 ) 

Ta có : 

\(\left(x+5\right)^2\ge0\forall x\) 

\(\left(y+3\right)^2\ge0\forall y\) 

\(\left(1\right)=0\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\)   

\(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\)         

\(\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)   

b) 

\(25x^2+10x+1+4y^2+4y+1=0\)     

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

Ta có : 

\(\left(5x+1\right)^2\ge0\forall x\)      

\(\left(2y+1\right)^2\ge0\forall y\)  

\(\left(1\right)=0\Leftrightarrow\hept{\begin{cases}\left(5x+1\right)^2=0\\\left(2y+1\right)^2=0\end{cases}}\)   

\(\hept{\begin{cases}5x+1=0\\2y+1=0\end{cases}}\)    

\(\hept{\begin{cases}x=\frac{-1}{5}\\y=\frac{-1}{2}\end{cases}}\)

x² + y² + 10x + 6y + 34 = 0

<=> ( x² + 10x + 25 ) + ( y² + 6y + 9 ) = 0

<=> ( x + 5 )² + ( y + 3 )² = 0

<=> \(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)

25x² + 4y² + 10x + 4y + 2 = 0

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

<=> ( 5x + 1 )² + ( 2y + 1 )² = 0

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

a) =2x - 3 =0

     x = 3/2

b) (5x -1)² = 0

     5x - 1 =  0

       x = 1/5

c) = ( x +3)² = 0

        x+3  = 0

         x = -3

d) =(13+y)(13-y) = 0

        y = 13; -13

e) xem lại đề bài này

thank bạn nhiều lắm

a ) ( 2 x - 3 ) ^ 2 = 0

=> 2 x - 3           = 0

     2 x                = 3

        x                = 1,5

b ) 25 x ^ 2  - 10 x + 1 = 0

  (  5 x ) ^ 2   -  2 . 5 x + 1 ^ 2 = 0

  ( 5 x - 1 ) ^  2                       = 0

  5 x - 1                                  = 0

  5x                                        = 1

    x                                        = 0,2

c ) 6 x + 9 = - x ^ 2

   6 x + 9 + x ^ 2 = 0

  x ^ 2 + 2 . x . 3 + 3 ^ 2 = 0

  ( x + 3 ) ^ 2                  = 0

  x + 3                            = 0

        x                            = -3

d ) 169 - y ^ 2 = 0

            y  ^ 2  = 169

            y ^ 2   = 13 ^ 2

=> y     = 13

Bài 1:Tìm x,y biết:

a)\(x^2-6x+y^2+10y+34\)

=>\(\left(x^2-2.x.3+3^2\right)+\left(y^2+2.y.5+5^2\right)=0\)

=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)

=>\(\left\{{}\begin{matrix}x-3=0\\y+5=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)

Còn ý b,c,d,e làm tương tự ý a.

Đáp án : D

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

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

b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

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

c, \(5x-2-25x^2+10x=0\)

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

= 1/5 nha

a)

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

\(\Rightarrow\left(x+2\right)^2=9=3^2\)

\(\Rightarrow x+2=\pm3\)

\(\Rightarrow x=-5;1\)

b)

\(25x^2-10x+1=0\)

\(\left(5x\right)^2-2\cdot5x+1^2=0\)

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

\(\Rightarrow5x+1=0\)

\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)

c)

\(x^2+14x+49=0\)

\(\Rightarrow x^2+2\cdot7x+7^2=0\)

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

\(\Rightarrow x=-7\)

d)

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)

\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)

\(\Rightarrow2x+255=0\)

\(\Rightarrow2x=-255\)

\(\Rightarrow x=\dfrac{-255}{2}\)

a.

\(\sqrt{4x^2+4x+1}-\sqrt{25x^2+10x+1}=0\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}-\sqrt{\left(5x+1\right)^2}=0\)

\(\Leftrightarrow2x+1-\left(5x+1\right)=0\)

\(\Leftrightarrow-3x=0\Leftrightarrow x=0\)

b.

\(\sqrt{x^4-16x^2+64}=\sqrt{25x^2+10x+1}\)

\(\Leftrightarrow\sqrt{\left(x^2-8\right)^2}=\sqrt{\left(5x+1\right)^2}\)

\(\Leftrightarrow x^2-8=5x+1\)

\(\Leftrightarrow x^2-5x+\dfrac{25}{4}=\dfrac{61}{4}\)

\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{61}{4}\)

............................

tương tự ..

c: \(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x+5}-1\right)=0\)

=>x-5=0 hoặc x+5=1

=>x=-4 hoặc x=5

d: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)

=>2x+3=0 hoặc 2x-3=4

=>x=7/2 hoặc x=-3/2

e: \(\Leftrightarrow\sqrt{x-2}\left(1-3\sqrt{x+2}\right)=0\)

=>x-2=0 hoặc 3 căn x+2=1

=>x=2 hoặc x+2=1/9

=>x=-17/9 hoặc x=2

1, \(3x\left(x-7\right)+2x-14=0\)

\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)

2, \(x^3+3x^2-\left(x+3\right)=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)

3, \(15x-5+6x^2-2x=0\)

\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)

\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)

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

4, \(5x-2-25x^2+10x=0\)

\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)

\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)

\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)

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

X= -1; X=6; X=2; X=3

SUY RA \(x^4+x^3-11x^3-11x^2+36x^2-36=0\)

          \(\Leftrightarrow x^3\left(x+1\right)-11x^2\left(x+1\right)+36\left(x-1\right)\left(x+1\right)=0\)

          \(\Leftrightarrow\left(x^3-11x^2+36x-36\right)\left(x+1\right)=0\)

           \(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x-2\right)\left(x+1\right)=0\)

           suy ra x=-1 hoặc x=6 hoặc x=3 hoặc x=2

mk làm hơi tắt nhưng vẫn dk k nha

a)  2x² - 98 = 0

     2x²        = 0 + 98

     2x²        = 98

       x²        = 98 : 2

       x²         = 49

       x          = \(\sqrt{49}\)

=>   x   = 7

Ta có : 2x² - 98 = 0

=> 2(x² - 49) = 0

Mà : 2 > 0

Nên x² - 49 = 0

=> x² = 49

=> x² = -7;7

2x²-98=0

=>2(x²-49)=0

=>x²-49=0

x²=49+0

x²=7²

x=7