1.(2x+3).(x-5)+2x(3-x)+x-10

=2x^2 -10x+3x-15+6x-2x^2+x-10

=2x-25

2.(-x-2)³+(2x-4).(x²+2x+4)-x².(x-6)

=-x^3+6x^2-12x-8+2x^3+4x^2+8x-4x^2+8x-16-x^3+6x^2

phần dưới là tìm x

\(\dfrac{5x+2}{x^2-4}+\dfrac{x-5}{x-2}=\dfrac{5x+2+x^2-3x-10}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+4}{x+2}\\ \left(x+4\right)^2-\left(x+3\right)\left(x-2\right)=-13\\ \Leftrightarrow x^2+8x+16-x^2+x+6=-13\\ \Leftrightarrow9x=-13-22=-35\\ \Leftrightarrow x=-\dfrac{35}{9}\)

a: \(\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)

\(=\dfrac{4-x^2-2x+2x^2+5-4x}{x-3}=\dfrac{x^2-6x+9}{x-3}\)

=(x-3)^2/(x-3)

=x-3

b: \(\dfrac{2}{x+2}+\dfrac{-4}{2-x}+\dfrac{5x+2}{4-x^2}\)

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

\(=\dfrac{2x-4-4x-8-5x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-7x-14}{\left(x-2\right)\left(x+2\right)}\)

=-7(x+2)/(x-2)(x+2)

=-7/(x-2)

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

a: \(=\dfrac{5}{2x^2y}+\dfrac{2}{3xy}-\dfrac{y}{x^3}\)

\(=\dfrac{5\cdot3\cdot x}{6x^3y}+\dfrac{2\cdot2\cdot x^2}{6x^3y}-\dfrac{6y^2}{6x^3y}\)

\(=\dfrac{15x+4x^2-6y^2}{6x^3y}\)

b: \(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)

c: \(=\dfrac{x^4-1-x^4+3x^2}{x^2-1}=\dfrac{3x^2-1}{x^2-1}\)

a: \(=2x^2-4x-6x+12-2x^2+10x=12\)

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

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

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

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

Bài 4:

a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)

\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)

\(=7x-16\)

b, = x² - 16 - ( x³ - 3³ ) : ( x - 3 )

= x² - 16 - \([\) ( x - 3 ) ( x² + 3x + 9 ) \(]\) : ( x - 3 )

= x² - 16 - ( x² + 3x + 9 )

= x² - 16 - x² - 3x - 9

= -25 - 3x

\(a,=2x^2-10x+3x-15-x^2+2x-1+7x-x^2=2x-16\\ b,=x^2-16-\left(x-3\right)\left(x^2+3x+9\right):\left(x-3\right)\\ =x^2-16-x^2-3x-9=-3x-25\)

Bài 2:

1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)

\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)

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

\(=x^3+8-2x^2+2=x^3-2x^2+10\)

\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)

\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)

\(=\left(-2y\right)^2+4\left(y+2\right)\)

\(=4y^2+4y+8\)

2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)

3: \(B=4y^2+4y+8\)

\(=4y^2+4y+1+7\)

\(=\left(2y+1\right)^2+7>=7>0\forall y\)

=>B luôn dương với mọi y

Bài 1:

5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)

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

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

6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)

\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)

\(=6x^2+23x-55-6x^2-84x-294\)

=-61x-349