a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))

\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)

\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)

\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)

\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)

\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)

b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

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

\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)

\(=1+\dfrac{3}{x}\)

\(=\dfrac{x}{x}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x}\)

a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)

\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)

b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)

a) \(\frac{3x}{2x+4}+\frac{x+3}{x^2-4}\)

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

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

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

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

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

\(ĐKXĐ:x\ne\pm2\)

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

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

\(=\left[\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right]:\left[\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right]\)

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

\(=\frac{2x}{\left(x+2\right)^2}:\frac{-x}{\left(x-2\right)\left(x+2\right)}=\frac{2x}{\left(x+2\right)^2}.\frac{-\left(x-2\right)\left(x+2\right)}{x}\)

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

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

\(\Leftrightarrow\left(\frac{2}{x+2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}+\frac{1}{2-x}\right)\)

\(\Leftrightarrow\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2+x+2}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{2x}{\left(x+2\right)^2}\cdot\frac{\left(x-2\right)\left(x+2\right)}{x+4}\)

\(\Leftrightarrow\frac{2x^2-4x}{\left(x+2\right)\left(x+4\right)}\)

Ta có \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\)

\(=\frac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x-2+x+2\right)\left(x-2-x-2\right)}{\left(x-2\right)^2\left(x+2\right)^2}:\frac{2x}{\left(x+2\right)\left(x-2\right)}\)

\(\frac{-4.2x}{\left(x+2\right)^2\left(x-2\right)^2}.\frac{\left(x+2\right)\left(x-2\right)}{2x}=\frac{-4}{\left(x+2\right)\left(x-2\right)}\)

khó quá làm sao mà trả lời đc

Vắt óc đi

tự đầu mình vắt óc mà suy nghĩ

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

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

\(=\frac{3x^2-6x^2-12x+24}{x^3+2x^2-4x-8}\)

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

\(=\frac{3x-6}{x+2}\)

\(\frac{x^2+4x+4}{1-x}.\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\)

\(=\frac{x^2+4x+4}{1-x}.\left[\frac{\left(1-x\right)^2}{3\left(x+2\right)^3}\right]\)

\(=\frac{x^4+2x^3-3x^2-4x+4}{-3x^4-15x^3-18x^2+12x+24}\)

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

\(=\frac{-x+1}{3x+6}\)

\(a,\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)

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

\(x\left(x+1\right)+x\left(x-3\right)=4x\)

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

\(2x^2-2x=4x\)

\(2x^2-2x-4x=0\)

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

\(2x=0\Leftrightarrow x=0\)

hoặc 

\(x-3=0\Leftrightarrow x=3\)

b) \(ĐKXĐ:x\ne\pm4\)

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)

\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)

\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)

\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)

\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)

\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)

\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )

Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)