a: Ta có: \(x^2-8x+20\)

\(=x^2-8x+16+4\)

\(=\left(x-4\right)^2+4>0\forall x\)

b: Ta có: \(-x^2+6x-19\)

\(=-\left(x^2-6x+19\right)\)

\(=-\left(x^2-6x+9+10\right)\)

\(=-\left(x-3\right)^2-10< 0\forall x\)

Sửa đề: \(A=3x^2-6x+4=3\left(x^2-2x+\dfrac{4}{3}\right)\)

\(A=3\left(x^2-2x+1+\dfrac{1}{3}\right)\)

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

\(A=3\left(x-1\right)^2+1>0\left(đpcm\right)\)

a) x² - 2x + 4x - 8 = 0

=> x.(x - 2) + 4.(x - 2) = 0

=> (x - 2).(x + 4) = 0

=> \(\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}}\)=> \(\orbr{\begin{cases}x=2\\x=-4\end{cases}}\)

b) x(x + 3) - 3x - 9 = 0

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

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

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

c) x² - 6x + 5 = 0

=> x² - x - 5x + 5 = 0

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

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

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

1/\(x^2-2x+4x-8=0\)

=>\(x\left(x-2\right)+4\left(x-2\right)=0\)

=>\(\left(x-4\right)\left(x-2\right)=0\)

=>\(\orbr{\begin{cases}x-4=0\\x-2=0\end{cases}}\)=>\(\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

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

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

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

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

3/\(x^2-6x+5=0\)

=>\(x^2-x-5x+5=0\)

=>\(x\left(x-1\right)-5\left(x-1\right)=0\)

=>\(\left(x-5\right)\left(x-1\right)=0\)

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

mình nhầm câu a, chỗ x-4 sửa thành x+4 và => x=-4 chứ ko phải là 4

\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=x+1-\left(x-9\right)\)

\(\Rightarrow6x^2+27x+4x+18-6x^2-x-12x-2=10\)

\(\Rightarrow18x+16=10\)

\(\Rightarrow18x=-6\)

\(\Rightarrow x=-\frac{6}{18}=-\frac{1}{3}\)

cam on 

\(2x^2+7x+3=0\)

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

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

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

đến đây tự làm

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

\(\Leftrightarrow\)\(\left(2x-1\right)^2+11=0\)vô lý

Vậy pt vô nghiệm

\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow\left(x^2+x^3-2x^2\right)+\left(x^3+x^2-2x\right)+\left(6x^2+6x-12\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=0\\x^2+x-2=0\end{matrix}\right.\)

* \(x^2+x+6=\left(x^2+x+\frac{1}{4}\right)+\frac{23}{4}=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\)

\(\Rightarrow x^2+x+6=0\) là vô lí

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

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

a) \(2x^3+x^2-4x-12\)

\(=2x^3-4x^2+5x^2-10x+6x-12\)

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

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

b) \(5x^2+6xy+y^2\)

\(=5x^2+5xy+xy+y^2\)

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

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