mình cần gấp mong các bạn giải giùm

c: \(E=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)

\(12+3y+4x+xy-12+2y+6x-x-5y\)

\(=9x+xy\)

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

12+3y+4x+xy-12+2y+6x-x-5y=

=9x+xy

=x(9+y)

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

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

chịu

\(A=\dfrac{x+12}{x-4}+\dfrac{1}{\sqrt{x}+2}-\dfrac{4}{\sqrt{x}-2}\)

\(=\dfrac{x+12+\sqrt{x}-2-4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+\sqrt{x}+10-4\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

x−3√x+2/(√x+2)(√x−2) giải chi tiết ra đi

(-3).8/8.6 rút gọn

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(B=\dfrac{6-7x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{2}{2-x}\)

\(=\dfrac{6-7x+3x-6+2x+4}{\left(x+2\right)\left(x-2\right)}\)

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

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

a: ĐKXĐ: x∉{2;-2;3}

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x^2-x-6}\)

\(=\left(\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\frac{16}{\left(x-3\right)\left(x+2\right)}\)

\(=\left(\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\frac{\left(x-3\right)\cdot\left(x+2\right)}{16}\)

\(=\frac{4x\left(x+2\right)-4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\cdot\frac{\left(x-3\right)\left(x+2\right)}{16}=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)}\cdot\frac{x-3}{16}\)

\(=\frac{-16\left(x-3\right)}{16\left(x+2\right)}=\frac{-x+3}{x+2}\)

b: A<0

=>\(\frac{-x+3}{x+2}<0\)

=>\(\frac{x-3}{x+2}>0\)

=>x>3 hoặc x<-2

c: A>=5

=>\(\frac{-x+3}{x+2}-5\ge0\)

=>\(\frac{-x+3-5x-10}{x+2}\ge0\)

=>\(\frac{-6x-7}{x+2}\ge0\)

=>\(\frac{6x+7}{x+2}\le0\)

=>-2<x<=-7/6