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a.\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)=\(\dfrac{x+1}{x\left(x+1\right)}\)-\(\dfrac{x}{x\left(x+1\right)}\)=\(\dfrac{x+1-x}{x\left(x+1\right)}\)=\(\dfrac{1}{x\left(x+1\right)}\)
b. Ta có:
\(\dfrac{1}{x\left(x+1\right)}\)= \(\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}\)=\(\dfrac{x+1}{x\left(x+1\right)}\)-\(\dfrac{x}{x\left(x+1\right)}\)=\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)
Ta lại có:
\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)=\(\dfrac{1}{x+1}\)-\(\dfrac{1}{x+2}\);
\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)=\(\dfrac{1}{x+2}\)-\(\dfrac{1}{x+3}\);
\(\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)=\(\dfrac{1}{x+3}\)-\(\dfrac{1}{x+4}\);
\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)=\(\dfrac{1}{x+4}\)-\(\dfrac{1}{x+5}\);
Do đó:
\(\dfrac{1}{x\left(x+1\right)}\)+\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)+\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)+\(\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)+\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)+\(\dfrac{1}{x+5}\) = \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\)+\(\dfrac{1}{x+1}\)-\(\dfrac{1}{x+2}\)+\(\dfrac{1}{x+2}\)-...... -\(\dfrac{1}{x+5}\)+\(\dfrac{1}{x+5}\)=\(\dfrac{1}{x}\)
Vậy tổng trên bằng \(\dfrac{1}{x}\)
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
a: =>1+3x-6=-x+3
=>3x-5=-x+3
=>4x=8
=>x=2(loại)
b: \(\Leftrightarrow\dfrac{3\left(x-3\right)+2\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{x-1}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
=>3x-9+2x-4=x-1
=>5x-13=x-1
=>4x=12
=>x=3(loại)
c: =>x^2-2x+4+x^3+8=12
=>x^3+x^2-2x=0
=>x(x^2+x-2)=0
=>x(x+2)(x-1)=0
=>x=0 hoặc x=1
Ta có \(\left(x-\dfrac{1}{x}\right):\left(x+\dfrac{1}{x}\right)=\dfrac{1}{2}\Leftrightarrow\dfrac{x^2-1}{x^2+1}=\dfrac{1}{2}\Leftrightarrow x^2=3\).
Do đó: \(\left(x^2-\dfrac{1}{x^2}\right):\left(x^2+\dfrac{1}{x^2}\right)=\dfrac{3-\dfrac{1}{3}}{3+\dfrac{1}{3}}=\dfrac{8}{10}=\dfrac{4}{5}\).
\(\left(\frac{1}{1+x}+\frac{2}{1+x^2}\right)\left(\frac{1-x}{x}\right)=\left(\frac{3+x^2+2x}{1+x+x^2+x^3}\right)\left(\frac{1-x}{x}\right)\)
\(\frac{3+x^2+2x}{1+x+x^2+x^3}.\frac{1-x}{x}=\frac{3+x^2+2x-3x-x^3-2x^2}{x+x^2+x^3+x^4}=\frac{3-x-x^2-x^3}{x+x^2+x^3+x^4}\)