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

( ĐKXĐ : x ≠ 1 ; x ≠ 2 )

\(\frac{1}{x^2+5x-6}=\frac{1}{x^2-x+6x-6}=\frac{1}{x\left(x-1\right)+6\left(x-1\right)}=\frac{1}{\left(x-1\right)\left(x+6\right)}\)

( ĐKXĐ : x ≠ 1 ; x ≠ -6 )

\(\frac{1}{-x^2+4x-3}=\frac{-1}{x^2-4x+3}=\frac{-1}{x^2-x-3x+3}=\frac{-1}{x\left(x-1\right)-3\left(x-1\right)}=\frac{-1}{\left(x-1\right)\left(x-3\right)}\)

( ĐKXĐ : x ≠ 1 ; x ≠ 3 )

MTC : ( x - 1 )( x - 2 )( x - 3 )( x + 6 )

=> \(\frac{-1}{\left(x-1\right)\left(x-2\right)}=\frac{-1\left(x+6\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}=\frac{-x^2-3x+18}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}\)

=>\(\frac{1}{\left(x-1\right)\left(x+6\right)}=\frac{1\left(x-2\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}=\frac{x^2-5x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}\)

=> \(\frac{-1}{\left(x-1\right)\left(x-3\right)}=\frac{-1\left(x-2\right)\left(x+6\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}=\frac{-x^2-4x+12}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}\)

\(\frac{1}{x^2-2x+1}=\frac{1}{\left(x-1\right)^2}\)( ĐKXĐ : x ≠ 1 )

\(\frac{2}{x^2+2x}=\frac{2}{x\left(x+2\right)}\)( ĐKXĐ : x ≠ 0 ; x ≠ -2 )

MTC : x( x + 2 )( x - 1 )²

Thực hiện phép tính

\(5x^2\left(4x^2-2x+5\right)\)

\(=20x^4-10x^3+25x^2\)

b) \(\left(6x^2-5\right)\left(2x+3\right)\)

\(=6x^2\left(2x+3\right)-5\left(2x+3\right)\)

\(=12x^3+18x^2-10x-15\)

a) x² + 7x + 12

= x² + 3x + 4x + 12

= ( x² + 3x ) + ( 4x + 12 )

= x( x + 3 ) + 4( x + 3 )

= ( x + 3 )( x + 4 )

b) x⁴ + 8x² + 16

= ( x² )² + 2.x².4 + 4²

= ( x² + 4 )²

c) x² + 26x + 25

= x² + x + 25x + 25

= ( x² + x ) + ( 25x + 25 )

= x( x + 1 ) + 25( x + 1 )

= ( x + 1 )( x + 25 )

a) \(x^2+7x+12\)

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

\(x\left(x+4\right)+3\left(x+4\right)\)

\(\left(x+4\right)\left(x+3\right)\)

b) \(x^4+8x^2+16\)

\(\left(x^2+4\right)^2\)

c) \(x^2+26x+25\)

\(x^2+x+25x+25\)

\(x\left(x+1\right)+25\left(x+1\right)\)

\(\left(x+25\right)\left(x+1\right)\)

\(2x^2+4y^2-4xy+x-4y=-\frac{5}{4}\)

\(\Leftrightarrow\text{​​}x^2-x+\frac{1}{4}+x^2-2x\left(2y-1\right)+4y^2-4y+1=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\x-2y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{4}\end{cases}}}\)

Ta có : x² - x + 1

= x² - x + 1/4 + 3/4

= ( x² - x + 1/4 ) + 3/4

= ( x - 1/2 )² + 3/4 ≥ 3/4 ∀ x

Dấu "=" xảy ra khi x = 1/2

=> GTNN của biểu thức = 3/4 <=> x = 1/2

\(x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge0\forall x\)

Dấu ''='' xãy ra <=> \(x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)

Vậy GTNN của biểu thức là 3/4 <=>  x = 1/2 

( x - 10 )² - x( x + 80 )

= x² - 20x + 100 - x² - 80x

= -100x + 100

Với x = 0, 98 = 98/100

=> Giá trị biểu thức = -100.98/100 + 100 = -98 + 100 = 2 

Ta có : 

\(\left(x-10\right)^2-x\left(x+80\right)=x^2-20x+100-x^2-80x=-100x+100=-100\left(x-1\right)\)

Thay x = 0,98 vào biểu thức trên ta được :

\(-100\left(0,98-1\right)\approx0,2\)