Tương tự mấy phần kia 

\(A=\frac{x+3}{x-2}+\frac{x+2}{3-x}+\frac{x+2}{x^2-5x+6}\)

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

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

\(=\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}=\frac{-3+x}{\left(x-2\right)\left(x-3\right)}=\frac{-1}{x-2}\)

\(\frac{1}{x-2}-\frac{1}{x+2}+\frac{4x-x^2}{4-x^2}\)

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

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

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

\(\frac{1}{x-2}-\frac{1}{x+2}+\frac{4x-x^2}{4-x^2}\)

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

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

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

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

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

\(\frac{2x^2-1}{x^2-xy}+\frac{1-2y^2}{x^2-xy}=\frac{2x^2-1+1-2y^2}{x^2-xy}\)

\(=\frac{2x^2-2y^2}{x^2-xy}=\frac{2\left(x^2-y^2\right)}{x\left(x-y\right)}\)

\(=\frac{2\left(x-y\right)\left(x+y\right)}{x\left(x-y\right)}=\frac{2\left(x+y\right)}{x}\)

Sửa đề : \(\frac{2x^2-1}{x^2-xy}+\frac{1-2y^2}{x^2-xy}\)

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

\(=\frac{2\left(x-y\right)\left(x+y\right)}{x\left(x-y\right)}=\frac{2\left(x+y\right)}{x}\)

a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)

b, \(x^3+x^2-9x-9=0\Leftrightarrow x^2\left(x+1\right)-9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-9\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=-1;\pm3\)

c, \(x^2-3x-10=0\Leftrightarrow x^2+2x-5x-10=0\)

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

a, \(20x^2y^3-15xy^2=5xy^2\left(4xy-3\right)\)

b, \(3x+3y-x^2-xy=3\left(x+y\right)-x\left(x+y\right)=\left(3-x\right)\left(x+y\right)\)

c, \(9-x^2-y^2+2xy=9-\left(x^2+y^2-2xy\right)\)

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

20x²y³ - 15xy² = 5xy²( 4xy - 3 )

3x + 3y - x² - xy = ( 3x + 3y ) - ( x² + xy ) = 3( x + y ) - x( x + y ) = ( x + y )( 3 - x )

9 - x² - y² + 2xy = 9 - ( x² - 2xy + y² ) = 3² - ( x - y )² = ( 3 - x + y )( 3 + x - y )

là gì?

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)0a

Đặt f(x) = 2x² + ax + 1

      g(x) = x - 3

f(x) chia g(x) dư 4 

=> f(x) - 4 chia hết cho g(x)

<=> 2x² + ax + 1 - 4 chia hết cho x - 3

<=> 2x² + ax - 3 chia hết cho x - 3

Áp dụng định lí Bézout ta có :

f(x) - 4 chia hết cho g(x) <=> f(3) - 4 = 0

<=> 18 + 3a - 3 = 0

<=> 3a + 15 = 0

<=> 3a = -15

<=> a = -5

Vậy a = -5