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\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+....+\dfrac{1}{\left(x+2017\right)\left(x+2018\right)}\\ =\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\\ =\dfrac{1}{x}-\dfrac{1}{x+2018}\\ =\dfrac{2018}{x\left(x+2018\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2016}-\dfrac{1}{x+2017}+\dfrac{1}{x+2017}-\dfrac{1}{x+2018}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2018}\)
\(=\dfrac{2018}{x\left(x+2018\right)}\)
a) \(\dfrac{2}{3x+9}-\dfrac{x-3}{3x^2+9x}\)
\(=\dfrac{2}{3\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)
\(=\dfrac{2x}{3x\left(x+3\right)}-\dfrac{x-3}{3x\left(x+3\right)}\)
\(=\dfrac{2x-x+3}{3x\left(x+3\right)}\)
\(=\dfrac{x+3}{3x\left(x+3\right)}\)
\(=\dfrac{1}{3x}\)
b) \(\dfrac{x^2+x}{5x^2-10x+5}:\dfrac{3x+3}{5x-5}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x^2-2x+1\right)}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}:\dfrac{3\left(x+1\right)}{5\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{5\left(x-1\right)^2}.\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\)
\(=\dfrac{x}{\left(x-1\right).3}\)
\(=\dfrac{x}{3x-3}\)
c) \(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+99}-\dfrac{1}{x+100}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+100}\)
\(=\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)
\(=\dfrac{x+100-x}{x\left(x+100\right)}\)
\(=\dfrac{100}{x\left(x+100\right)}\)
a) Đk : \(x\ne0;\ne1\)
\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)
\(\Rightarrow\dfrac{x^2+3x}{x\left(x+1\right)}+\dfrac{x^2-x-2}{x\left(x+1\right)}-\dfrac{2x^2+2x-2}{x\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{x^2+3x+x^2-x-2-2x^2-2x+2}{x\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{0}{x-1}=0\)
=> Phương trình có vô số nghiệm x
b) Đk : \(x\ne2;x\ne3\)
\(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)
\(\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}\)
=0
\(\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0\)
=> Phương trình vô nghiệm
c)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{-1}{x^4+x^2+1}=0\)
=> PTVN
d) Thôi tự làm đi, câu này dễ :Vvv
e)
\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)\)=40
\(\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40\)
\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt
\(x^2+6x+7=t\)
Phương trình tương đương
\(\left(t-1\right)\left(t+1\right)=40\)
\(t^2=41\)
\(\)\(t=\pm\sqrt{41}\)
Thay vào tìm x.
\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+13\right)}=\dfrac{12}{13}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+13}=\dfrac{12}{13}\)
\(\Leftrightarrow12\left(x+1\right)\left(x+13\right)=13\left(x+13\right)-13\left(x+1\right)=156\)
\(\Leftrightarrow\left(x+1\right)\left(x+13\right)=13\)
\(\Leftrightarrow x^2+14x=0\)
=>x=0 hoặc x=-14