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\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..................+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+..................+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.......+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+......+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{99}-\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+......+\dfrac{1}{50}\right)\)
\(=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+.......+\dfrac{1}{100}\)
\(\Leftrightarrow x\left(2x^2+10x-x-5\right)-\left(2x^3+9x^2+x+4.5\right)=3.5\)
\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4.5=3.5\)
=>-6x=8
hay x=-4/3
\(A-1=\frac{1}{1.2}+\frac{1}{2.3}..+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}\)\(=\frac{99}{100}\)
\(A=1+\frac{99}{100}=\frac{199}{100}\)
\(1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=2-\frac{1}{100}\)
\(=\frac{199}{100}\)
Gọi biểu thức là A
A=1+1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1-1/2+1/2-1/3+1/3-1/4+...+/198-1/99+1/99-1/100
A-1=1-1/100
A-1=99/100
A=99/100+1
A=199/100
\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)
\(\Rightarrow\left(2x^2-x\right)\left(x+5\right)-2x^3-9x^2-x-4,5=3,5\)
\(\Rightarrow2x^3+10x^2-x^2-5x-2x^3-9x^2-x-4,5=3,5\)
\(\Rightarrow-5x-4,5=3,5\)
\(\Rightarrow-5x=8\)
\(\Rightarrow x=-\dfrac{8}{5}\)
thoyy,tui xin chịu ạ
=1/4+1/5-1/5+...-1/39+1/40
=1/4+1/19
=.....(tự tính nka)