đk : x khác -2 ; 2 ; -1 

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

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

\(\Rightarrow P=\left(\dfrac{x+2}{\text{​​}\left(x-2\right)\left(x+2\right)}-\dfrac{4}{\left(x-2\right)\left(x+2\right)}\right).\left(\dfrac{x+1}{x+1}+\dfrac{1}{x+1}\right)\)

\(\Rightarrow P=\dfrac{x+2-4}{\text{​​}\left(x-2\right)\left(x+2\right)}.\dfrac{x+1+1}{x+1}\)

\(\Rightarrow P=\dfrac{x-2}{\text{​​}\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{x+1}\)

\(\Rightarrow P=\dfrac{1}{x+1}\)

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

đk : xkhác -1 ; 1 ; 1/2

\(A=\dfrac{3-3x+x+1-8}{\left(1-x\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}=\dfrac{-2x-4}{\left(1-x\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)

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

\(=\left(\dfrac{3}{x+1}-\dfrac{1}{x-1}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x^2-1}{1-2x}\)

\(=\dfrac{3x-3-x-1+8}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-2x+1}\)

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

ĐKXĐ:\(x\ne\pm1,x\ne\dfrac{1}{2}\)

\(A=\left(\dfrac{3}{x+1}+\dfrac{1}{1-x}-\dfrac{8}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

\(\Rightarrow A=\left(\dfrac{3}{x+1}-\dfrac{1}{x-1}+\dfrac{8}{x^2-1}\right).\dfrac{x^2-1}{1-2x}\)

\(\Rightarrow A=\left(\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{8}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{x^2-1}{1-2x}\)

\(\Rightarrow A=\dfrac{3x-3-x-1+8}{\left(x-1\right)\left(x+1\right)}.\dfrac{x^2-1}{1-2x}\)

\(\Rightarrow A=\dfrac{2x+4}{x^2-1}.\dfrac{x^2-1}{1-2x}\)

\(\Rightarrow A=\dfrac{2x+4}{1-2x}\)

giups vs, chỉ cần điền thì quá khứ và quá khứ hoàn thành tiếp diễn thôi. Dễ mà, giúp mik nhanh vs

1 had/had been decorating

2 drove/had been beginning

3 hadn't been living/decided

4 Had you been waiting/arrived

5 had been saving/didn't have

6 had been expecting/pulled

P/s:Tưởng cấp 3 mới học phần này thôi chứ? :v

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

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

\(\Rightarrow P=\dfrac{2x^2-6x+x^2+3x-3x^2-3}{\left(x+3\right)\left(x-3\right)}:\dfrac{2x-2-x+3}{x-3}\)

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

\(\Rightarrow P=\dfrac{-3\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}.\dfrac{x-3}{x+1}\)

\(\Rightarrow P=\dfrac{-3}{x+3}\)

Ta có: \(x^2< 3x\)

\(\Leftrightarrow x^2-3x< 0\)

\(\Leftrightarrow x\left(x-3\right)< 0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x-3< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x-3>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x< 3\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x>3\end{matrix}\right.\left(loại\right)\end{matrix}\right.\Leftrightarrow0< x< 3\)

Vậy: S={x|0<x<3}

đăng từ từ, ok nhìu thế này ai rỗi mà giải

a>

27-(5-/ p/)=31

5- /p /=27-31

5-/p/=-4

/p/ =5 - (-4)

/ p /=5+4

/p/=9

vậy  p =9 hoặc p =-9

b>

-13-(6-/p-1/)_=24

6-/p+1/=-13-24

6-/p+1/=-37

/p+1/=6-(-37)

/p+1)=6+37

/p+1/=43

vậy p + 1 = 37 hoặc p + 1 = - 37

      p       =37-1  ;    p       =-37 -1

       p     =36   ;        p    =-37 +     ( -1)

                              p      = -38

vậy  p     = 36 hoặc -38

k mk nha

a) \(\left(\frac{1}{2}x+y\right)\left(\frac{1}{2}x+y\right)\)

\(=\frac{1}{2}x.\left(\frac{1}{2}x+y\right)+y.\left(\frac{1}{2}x+y\right)\)

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

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

b) \(\left(x^3-2x^2+x\right).\left(5-x\right)\)

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

\(=5x^3-x^4-10x^2+2x^3+5x-x^2\)

\(=7x^3-x^4-11x^2+5x\)

\(y=-4\\ \Rightarrow-4=3x^2-7\\ \Rightarrow3x^2=3\\ \Rightarrow x^2=1\\ \Rightarrow x=\pm1\)

\(y=5\\ \Rightarrow5=3x^2-7\\ \Rightarrow3x^2=12\\ \Rightarrow x^2=4\\ \Rightarrow x=\pm2\)

\(y=-6\dfrac{2}{3}\\ \Rightarrow-6\dfrac{2}{3}=3x^2-7\\ \Rightarrow3x^2=\dfrac{1}{3}\\ \Rightarrow x^2=\dfrac{1}{9}\\ \Rightarrow x=\pm\dfrac{1}{3}\)