\(A=\left(\dfrac{4}{x-2}-\dfrac{3}{x+2}\right):\dfrac{x+14}{x^2}\\ =\left(\dfrac{4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x^2}{x+14}\\ =\dfrac{4x+8-3x+6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2}{x+14}\\ =\dfrac{x+14}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^2}{x+14}\\=\dfrac{x^2}{x^2-4}\)

Tại `x=-3` Ta có :

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

làm khuyến mại 1 câu;

a) = 12x² -12x² +20x -10x +17 =0

10x = -17

x = -17/10

x/2 - ( 3x/5 - 13/5 ) = -( 7/5 + 7/10x )

Ta có: \(x=\frac{-1-\sqrt5}{2}\)

=>\(x^2=\frac{\left(-1-\sqrt5\right)^2}{2^2}=\frac{6+2\sqrt5}{4}=\frac{3+\sqrt5}{2}\)

=>\(x^4=\left(\frac{3+\sqrt5}{2}\right)^2=\frac{14+6\sqrt5}{4}=\frac{7+3\sqrt5}{2}\)

\(x^2\cdot x=\frac{\left(3+\sqrt5\right)\left(-\sqrt5-1\right)}{4}=\frac{-\left(3+\sqrt5\right)\left(\sqrt5+1\right)}{4}=\frac{-\left(3\sqrt5+3+5+\sqrt5\right)}{4}=-\sqrt5-2\)

=>\(x^3=-\sqrt5-2\)

\(x^5=x^3\cdot x^2=\left(-\sqrt5-2\right)\cdot\frac{3+\sqrt5}{2}\)

\(=-\frac12\left(2+\sqrt5\right)\left(3+\sqrt5\right)=-\frac12\left(6+2\sqrt5+3\sqrt5+5\right)=-\frac12\left(11+5\sqrt5\right)\)

\(4x^5+4x^4-5x^3+2x-2\)

\(=4\cdot\frac{-1}{2}\left(11+5\sqrt5\right)+4\cdot\frac{7+3\sqrt5}{2}-5\left(-\sqrt5-2\right)+2\cdot\frac{-1-\sqrt5}{2}\) -2

\(=-2\left(11+5\sqrt5\right)+2\left(7+3\sqrt5\right)+5\left(\sqrt5+2\right)-\sqrt5-1-2\)

\(=-22-10\sqrt5+14+6\sqrt5+5\sqrt5+10-\sqrt5-3=-1\)

Ta có: \(B=2\cdot\left(4x^5+4x^4-5x^3+2x-2\right)^{2021}+2022\)

=-2+2022

=2020

tham khảo:

\(P=\)\(\dfrac{8}{x^2+4x}\)+\(\dfrac{5}{x+4}\) - \(\dfrac{2}{x}\)

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

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

\(=\) \(\dfrac{3}{x\left(x+4\right)}\)

\(=\) \(\dfrac{3}{x+4}\)

Khi \(x =\) \(\dfrac{1}{2}\) thì \(P=\) \(\dfrac{3}{\dfrac{1}{2}+4}\) \(=\) \(\dfrac{2}{3}\)

 C.ơn :v