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

\(\Leftrightarrow\)\(3x^2-6x+3-3x^2+15x=1\)

\(\Leftrightarrow\)\(9x=-2\)

\(\Leftrightarrow\)\(x=-\frac{2}{9}\)

Vậy...

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

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)

Vậy...

        \(x^2-2x+1=25\)

\(\Leftrightarrow\)\(\left(x+1\right)^2=25\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=5\\x+1=-5\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=4\\x=-6\end{cases}}\)

Vậy...

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

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

\(\Leftrightarrow\)\(3x^2-6x+3-3x^2+15x=1\)

\(\Leftrightarrow\)\(9x=-2\)

\(\Leftrightarrow\)\(x=-\frac{2}{9}\)

Vậy....

xin lỗi nhé, câu a mình làm sai:

a)  \(x^2-2x+1=25\)

\(\Leftrightarrow\)\(\left(x-1\right)^2=25\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

Vậy....

     \(x^2-2x+1=25\)

\(\Leftrightarrow\)\(\left(x-1\right)^2=25\)

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)

Vậy....

b)  \(3\left(x-1\right)^2-3x\left(x-5\right)=1\)

\(\Leftrightarrow\)\(3x^2-6x+3-3x^2+15x=1\)

\(\Leftrightarrow\)\(9x=-2\)

\(\Leftrightarrow\)\(x=-\frac{2}{9}\)

Vậy...

Cam on ban <3

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

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

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

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

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

khó phết

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

\(\left(2-y\right)\left(x+1\right)=7\)

⇒ (x+1) và (2-y) ϵ {-1;1;-7;7}

⇒ (x;y) ϵ {(-2;5);(0;-5);(-8;3);(6;1)}

\(\left(2x+1\right)\left(x-5\right)=2x^2-9x-5=2\left(x^2-\frac{9}{2}x+\frac{81}{16}\right)-\frac{121}{8}=2\left(x-\frac{9}{4}\right)^2-\frac{121}{8}\ge-\frac{121}{8} \)

Vậy GTNN của biểu thức là \(-\frac{121}{8}\)khi x = \(\frac{9}{4}\)

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

\(4x^2-20x+25-4x^2-x+8x+2=0\)

\(-13x+27=0\)

\(x=\frac{27}{13}\)

25 – 2x = 16 – 3x

-2x + 3x = 16 - 25

x = -9

Vậy x = -9

x - (47 - 22 ) = 5 + (10-4x)

x - 47 + 22    = 5 + 10 - 4x

x + 4x = 5 + 10 - 22 + 47

5x       = 40

x         = 40 : 5

x         = 8

Vậy x = 8.

~ HOK TỐT ~

\(\text{25 – 2x = 16 – 3x}\)

\(2x+3x=16-25\)

\(5x=-9\)

\(\Rightarrow x=\frac{-9}{5}\)

\(\text{d) x – (47 – 22) = 5+ (10 – 4x)}\)

\(x-47+22=5+10-4x\)

\(x+4x=5+10-22+47\)

\(5x=40\)

\(\Rightarrow x=8\)

học tốt

\(25-2x=16-3x\)

\(\Leftrightarrow25-2x-16=-3x\)

\(\Leftrightarrow-2x+9=-3x\)

\(\Leftrightarrow-3x+2x=9\)

\(\Leftrightarrow-x=9\Leftrightarrow x=9\)

Vậy ...

\(x-\left(47-22\right)=5+\left(10-4x\right)\)

\(\Leftrightarrow x-47+22=5+10-4x\)

\(\Leftrightarrow x-25=5+10-4x\)

\(\Leftrightarrow x-25=-4x+15\)

\(\Leftrightarrow x=-4x+15+25\)

\(\Leftrightarrow x=-4x+40\)

\(\Leftrightarrow x+4x=40\Leftrightarrow5x=40\Leftrightarrow x=8\)

Vậy ...