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

b. Với \(5x^2+3x=0\Leftrightarrow x\left(5x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\) nhưng mà ở trên ta cần có điều kiện x#0 nên

\(x=-\frac{3}{5}\Rightarrow A=\frac{3}{1-3\times\left(-\frac{3}{5}\right)}=\frac{15}{14}\)

c.\(A=\frac{x}{x-1}=\frac{3}{1-3x}\Leftrightarrow x-3x^2=3x-3\Leftrightarrow3x^2+2x-3=0\Leftrightarrow x=\frac{-1\pm\sqrt{10}}{3}\)

d.\(\frac{6}{A}=2\times\left(1-3x\right)\) nguyên nên \(1-3x=-\frac{k}{2}\Leftrightarrow x=\frac{k+2}{6}\) với k là số nguyên 

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

a: ĐKXĐ: \(x\notin\left\{0;1;2;3;4;5\right\}\)

b: \(P=\dfrac{1}{x^2-x}+\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}\)

\(=\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{-1}{x}+\dfrac{1}{x-1}-\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-4}-\dfrac{1}{x-4}+\dfrac{1}{x-5}\)

\(=\dfrac{1}{x-5}-\dfrac{1}{x}\)

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

c: \(x^3-x^2+2=0\)

=>\(x^3+x^2-2x^2+2=0\)

=>\(x^2\cdot\left(x+1\right)-2\left(x-1\right)\left(x+1\right)=0\)

=>\(\left(x+1\right)\left(x^2-2x+2\right)=0\)

=>x+1=0

=>x=-1

Khi x=-1 thì \(P=\dfrac{5}{\left(-1\right)\left(-1-5\right)}=\dfrac{5}{\left(-1\right)\cdot\left(-6\right)}=\dfrac{5}{6}\)

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

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

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

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

a: \(A=4x-3x^2+20-15x-9x^2-12x-4+\left(2x+1\right)^3-\left(8x^3-1\right)\)

\(=-12x^2-23x+16+8x^3+12x^2+6x+1-8x^3+1\)

\(=-17x+18\)

a, ĐK : \(x\ne1;2;3;4;5\)

b, \(\dfrac{1}{x\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}\)

\(=\dfrac{1}{x}-\dfrac{1}{x-1}+\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}\)

\(=\dfrac{1}{x}-\dfrac{1}{x-5}=\dfrac{x-5-x}{x\left(x-5\right)}=\dfrac{-5}{x\left(x-5\right)}\)

\(Bài.44:\\ a,3x-7=0\\ \Leftrightarrow3x=7\\ \Leftrightarrow x=\dfrac{7}{3}\\ b.2x^2+9=0\\ \Leftrightarrow x^2=-\dfrac{9}{2}\left(vô.lí\right)\\ \Rightarrow Không.có.x.thoả.mãn\)

43:

a: \(A=2x\left(x^2-2x-3\right)-6x^2+5x-1+9x^2+3x+3\)

\(=2x^3-4x^2-6x+3x^2+8x+2\)

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

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

Thương là x^2+1

Dư là 3

c: A chia hết cho 2x-1

=>3 chia hết cho 2x-1

=>2x-1 thuộc {1;-1;3;-3}

=>x thuộc {1;0;2;-1}