Bài 6: Phép trừ các phân thức đại số

\(=\dfrac{18}{\left(x-3\right)^2\cdot\left(x+3\right)}-\dfrac{3}{\left(x-3\right)^2}-\dfrac{x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{18-3x-9-x^2+3x}{\left(x-3\right)^2\cdot\left(x+3\right)}\)

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

Sửa đề: \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{4-9x^2}\)

\(=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x-2}\)

\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

\(=\dfrac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}\)

\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}\)
\(=\dfrac{-3}{x-3}\)

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

\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1\cdot x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{\left(x-3\right)\cdot3}{3x\left(x+3\right)}-\dfrac{x^2}{3x\left(x+3\right)}\right)\)

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}\)

\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)

\(=\dfrac{\left(9+x^2-3x\right)\cdot3x\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)\cdot\left(3x-9-x^2\right)}\)

\(=\dfrac{\left(9+x^2-3x\right)\cdot3x\left(x+3\right)}{-x\left(x-3\right)\left(x+3\right)\cdot\left(-3x+9+x^2\right)}\)

\(=\dfrac{3x}{-x\left(x-3\right)}\)

\(=\dfrac{3}{-\left(x-3\right)}\)

\(=\dfrac{-3}{x-3}\)

1,x(x+y)^2-y(x+y)^2 

`= x(x^2 + 2xy + y^2) -y(x^2+2xy+y^2) `

` = x^3 + 2x^2y + xy^2 - x^2y - 2xy^2 - y^3 ` 

`x(x+y)^2 - y(x+y)^2`

`= (x-y)(x+y)^2`

`= (x^2-y^2)(x+y)`

`= x^3 - xy^2 + x^2y - y^3`

không nhé

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

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

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

\(\Leftrightarrow\)\(\dfrac{x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x\left(x+2\right)}\)

\(\Leftrightarrow\dfrac{x^2+4x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{x\left(x+2\right)\left(x-2\right)}\)

\(\Leftrightarrow\dfrac{x^2+4x-x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+3x-2}{x\left(x-2\right)\left(x+2\right)}\)

\(a,5x^2y-10xy^2=5xy\left(x-2y\right)\\ b,x^2+2xy+y^2-5x-5y=\left(x+y\right)^2-5\left(x+y\right)=\left(x+y\right)\left(x+y-5\right)\\ c,x^2-6x+8=\left(x^2-2x\right)-\left(4x-8\right)=x\left(x-2\right)-4\left(x-2\right)=\left(x-2\right)\left(x-4\right)\\ d,5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\)

\(\dfrac{-5x+7-2x+3-3x+4}{2-x}=\dfrac{-10x+14}{2-x}=\dfrac{-2\left(5x-7\right)}{2-x}=\dfrac{2\left(5x-7\right)}{x-2}.\)

\(=\dfrac{3}{2\left(x+2\right)}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}=\dfrac{3x-6-2x}{2\left(x+2\right)\left(x-2\right)}=\dfrac{x-6}{2\left(x+2\right)\left(x-2\right)}\)

Bài 5: 

Xét ΔBAC có 

FG//AC

nên \(\dfrac{FG}{AC}=\dfrac{BG}{BC}=\dfrac{1}{2}\)

hay AC=16(m)