a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)

=3x+4

b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)

\(=\dfrac{x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=x+3\)

c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)

d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)

=7x+1

e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)

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

f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)

g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)

Đề lỗi rồi. Bạn xem lại

a) ĐKXĐ: x khác 0

\(x+\dfrac{5}{x}>0\)

\(\Leftrightarrow x^2+5>0\) ( luôn đúng)

Vậy bất pt vô số nghiệm ( loại x = 0)

d)

\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)

\(\Leftrightarrow2x+2-4x+4>-15\)

\(\Leftrightarrow-2x>-21\)

\(\Leftrightarrow x< \dfrac{21}{2}\)

Vậy....................

a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)

Mà \(x^2+5>0\)

\(\Rightarrow x>0\)

d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)

\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)

\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)

\(\Leftrightarrow-x>-\dfrac{21}{2}\)

\(\Leftrightarrow x< \dfrac{21}{2}\)

c)

\(\left(x^2-2x\right)\left(x^3-3x^2-18x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^3-6x^2+3x^2-18x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left[\left(x^3-6x^2\right)+\left(3x^2-18x\right)\right]=0\)

\(\Leftrightarrow x\left(x-2\right)\left[x^2\left(x-6\right)+3x\left(x-6\right)\right]=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x-6\right)\left(x^2+3x\right)=0\)

\(\Leftrightarrow x^2\left(x-6\right)\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\\x=2\\x=-3\end{matrix}\right.\)

Mọi người giúp mình nhanh nhé ! Mình đang cần gấp

b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)

\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)

\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)

\(\Leftrightarrow3x\left(x+4\right)=0\)

=>x=0(nhận) hoặc x=-4(loại)

\(a.7+2x=22-3x\\\Leftrightarrow 5x=15\\ \Leftrightarrow x=3\)

Vậy nghiệm của phương trình trên là \(3\)

\(b.8x-3=5x+12\\ \Leftrightarrow3x=15\\\Leftrightarrow x=5\)

Vậy nghiệm của phương trình trên là \(5\)

\(c.x-12+4x=25+2x-1\\ \Leftrightarrow5x-2x=12+24\\ \Leftrightarrow3x=36\\\Leftrightarrow x=12\)

Vậy nghiệm của phương trình trên là \(12\)

\(d.x+2x+3x-19=3x+5\\\Leftrightarrow 6x-3x=19+5\\\Leftrightarrow 3x=24\\ \Leftrightarrow x=8\)

Vậy nghiệm của phương trình trên là \(8\)

\(e.7-\left(2x+4\right)=-\left(x+4\right)\\ \Leftrightarrow7-2x-4=-x-4\\ \Leftrightarrow-2x+x=-3-4\\ \Leftrightarrow-x=-7\\ \Leftrightarrow x=7\)

Vậy nghiệm của phương trình trên là \(7\)

\(f.\left(x-1\right)-\left(2x-1\right)=9-x\\ \Leftrightarrow x-1-2x+1=9-x\\ \Leftrightarrow-x+x=9\\ \Leftrightarrow0x=9\)

\(\Rightarrow\) Vô nghiệm

\(a.\frac{4x-3}{x-5}=\frac{29}{3}\\ \Leftrightarrow\frac{3\left(4x-3\right)}{3\left(x-5\right)}=\frac{29\left(x-5\right)}{3\left(x-5\right)}\\ \Leftrightarrow3\left(4x-3\right)=29\left(x-5\right)\\ \Leftrightarrow3\left(4x-3\right)-29\left(x-5\right)=0\\ \Leftrightarrow12x-9-29x+145=0\\ \Leftrightarrow-17x+136=0\\ \Leftrightarrow-17x=-136\\ \Leftrightarrow x=\frac{-136}{-17}=8\)

\(b.\frac{2x-1}{5-3x}=2\\ \Leftrightarrow\frac{2x-1}{5-3x}=\frac{4}{2}\\ \Leftrightarrow\frac{2\left(2x-1\right)}{2\left(5-3x\right)}=\frac{4\left(5-3x\right)}{2\left(5-3x\right)}\\ \Leftrightarrow2\left(2x-1\right)=4\left(5-3x\right)\\ \Leftrightarrow2\left(2x-1\right)-4\left(5-3x\right)=0\\ \Leftrightarrow4x-2-20+12x=0\\ \Leftrightarrow16x-22=0\\ \Leftrightarrow16x=22\\ \Leftrightarrow x=\frac{22}{16}=\frac{11}{8}\)

\(c.\frac{4x-5}{x-1}=\frac{2+x}{x-1}\\ \Leftrightarrow4x-5=2+x\\ \Leftrightarrow4x-5-2-x=0\\ \Leftrightarrow3x-7=0\\ \Leftrightarrow3x=7\\ \Leftrightarrow x=\frac{7}{3}\)

\(d.\frac{7}{x+2}=\frac{3}{x-5}\\ \Leftrightarrow\frac{7\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-5\right)}\\ \Leftrightarrow7\left(x-5\right)=3\left(x+2\right)\\ \Leftrightarrow7\left(x-5\right)-3\left(x+2\right)=0\\ \Leftrightarrow7x-35-3x-6=0\\ \Leftrightarrow4x-41=0\\ \Leftrightarrow4x=41\\ \Leftrightarrow x=\frac{41}{4}\)

\(e.\frac{2x+5}{2x}-\frac{x}{x+5}=0\\ \Leftrightarrow\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{x.2x}{2x\left(x+5\right)}=0\\ \Leftrightarrow\left(2x+5\right)\left(x+5\right)-2x^2=0\\ \Leftrightarrow2x^2+10x+5x+25-2x^2=0\\ \Leftrightarrow15x+25=0\\ \Leftrightarrow15x=-25\\ \Leftrightarrow x=\frac{-25}{15}=\frac{-5}{3}\)

\(f.\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\\\Leftrightarrow\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{\left(10x-4\right).2\left(11x-4\right)}{9.2\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\\ \Leftrightarrow18\left(12x+1\right)+\left(10x-4\right).2\left(11x-4\right)=\left(20x+17\right)\left(11x-4\right)\\ \Leftrightarrow220x^2+48x+50=220x^2+107x-68\\ \Leftrightarrow48x+50=107x-68\\ \Leftrightarrow48x-107x=-68-50\\ \Leftrightarrow59x=-118\\ \Leftrightarrow x=-2\)