\(\dfrac{\left(x^2+6x+9\right)}{1-x}.\)\(\dfrac{\left(x-1^3\right)}{2\left(x+3\right)^3}\)

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

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

\(=\dfrac{x^2-2x+1}{-2x-6}\)

\(b,\left(3x+1\right)^2-\left(2x+1\right)^2\\ =\left[\left(3x+1\right)+\left(2x+1\right)\right]\left[\left(3x+1\right)-\left(2x+1\right)\right]\)

\(=\left(3x+1+2x+1\right)\left(3x+1-2x-1\right)\\ =x\left(5x+2\right)\)

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

\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\)

\(=\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-2\right)+x-14}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2+x+2x+2+x^2-2x-2x+4+x-14}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{2x^2-8}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{2\left(x^2-4\right)}{\left(x+2\right)\left(x-2\right)}\\ =\dfrac{2\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\\ =2\)

\(a,10x\left(x-y\right)-8\left(y-x\right)\\ =10x\left(x-y\right)+8\left(x-y\right)\\ =\left(x-y\right)\left(10x+8\right)\\ =2\left(5x+4\right)\left(x-y\right)\)

\(3^{x+3}-6.3^{x+1}=81\\ 3^x.3^3-6.3^x.3=81\\ 3^x\left(27-6.3\right)=81\\ 3^x.9=81\\ 3^x=9\\ 3^x=3^2\)

\(\Rightarrow x=2\)

a, Vì HE ⊥ AB ; FA ⊥ AB => HE // FA (từ ⊥ đến // )

+, EA ⊥ AC ; HF ⊥ AC => EA // HF (từ ⊥ đến // )

Xét tứ giác AEHF có: HE // FA (cmt) ; EA // HF (cmt)

=> Tứ giác AEHF là hình bình hành (dhnb)

 mà \(\hat{EAF} =90^0\)

=> Tứ giác AEHF là hình chữ nhật

=> AH = EF

b, Vì AEHF là hình chữ nhật (cmt)

=> EH//AF;  EH = AF mà AF= FK (gt)

=> EH = FK

+, Xét tứ giác EHKF có: EH = FK (cmt)

                                 EH // FK (do EH // AF; K ∈ AF)

=> Tứ giác EHKF là hình bình hành (dhnb)

\(\rightarrow\) \(B.\dfrac{6x}{15y}\)

\(\text{Giải thích:}\)

\(\dfrac{2x}{5y}=\dfrac{2x.3}{5y.3}=\dfrac{6x}{15y}\)

\(\dfrac{x^2+2x}{2x+12}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^2+2x}{2\left(x+6\right)}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x\left(x^2+2x\right)+2\left(x+6\right)\left(x-6\right)+108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+2x^2+2\left(x^2-36\right)+108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-72+108-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+4x^2+36-6x}{2x\left(x+6\right)}\)

\(=\dfrac{x^3+6x^2-2x^2-12x+6x+36}{2x\left(x+6\right)}\)

\(=\dfrac{\left(x^3+6x^2\right)+\left(-2x^2-12x\right)+\left(6x+36\right)}{2x\left(x+6\right)}\)

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

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

\(=\dfrac{x^2-2x+6}{2x}\)

\(a,2x\left(3x^2-5x+3\right)\\ =2x.3x^2-2x.5x+2x.3\\ =6x^3-10x+6x\)

\(b,\left(-2x-1\right).\left(x^2+5x-3\right)-\left(x-1\right)^3\\ =\left(-2x\right).x^2-2x.5x+2x.3-x^2-5x+3-\left(x^3-3x^2+3x-1\right)\\ =-2x^3-10x^2+6x-x^2-5x+3-x^3+3x^2-3x+1\\ =\left(-2x^3-x^3\right)+\left(-10x^2-x^2+3x^2\right)+\left(6x-5x-3x\right)+\left(3+1\right)\\ =-3x^3-8x^2-2x+4\)

\(A=\dfrac{x^2-2}{x^3+27}\)

A xác định 

\(\Leftrightarrow x^3+27\ne0\\ \Leftrightarrow x^3\ne-27\\ \Leftrightarrow x^3\ne\left(-3\right)^3\\ \Leftrightarrow x\ne-3\)