Phân thức đại số - Hỏi đáp

Từ gt \(\Leftrightarrow2A=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+...+\frac{2}{\left(x+2017\right)\left(x+2019\right)}\)

\(\Leftrightarrow2A=\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+....+\frac{1}{x+2017}-\frac{1}{x+2019}\)

\(\Leftrightarrow2A=\frac{1}{x+1}-\frac{1}{x+2019}\)

Với x = 3 thì :

\(2A=\frac{1}{4}-\frac{1}{2022}=\frac{1009}{4044}\)

\(\Rightarrow A=\frac{1009}{8088}\)

Chúc bạn học tốt !

Ta có :

\(a^3+b^3+c^3-3abc\)

\(=a^3+b^3+c^3+3a^2b+3ab^2-3a^2b-3ab^2-3abc\)

\(=\left[\left(a^3+3a^2b+3ab^2+b^3\right)+c^3\right]-\left(3a^2b+3ab^2+3abc\right)\)

\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)

\(=\left\{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\right\}-3ab\left(a+b+c\right)\)

\(=\left[\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)

\(\Rightarrow P=\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(=a+b+c\)

\(=2014\)

Vậy P = 2014

\(VT=\dfrac{b-c}{\left(a-b\right)\left(a-c\right)}+\dfrac{c-a}{\left(b-c\right)\left(b-a\right)}+\dfrac{a-b}{\left(c-a\right)\left(c-b\right)}\)

\(VT=\dfrac{b-a+a-c}{\left(a-b\right)\left(a-c\right)}+\dfrac{c-b+b-a}{\left(b-c\right)\left(b-a\right)}+\dfrac{a-c+c-b}{\left(c-a\right)\left(c-b\right)}\)

\(VT=\dfrac{-1}{a-c}+\dfrac{1}{a-b}+\dfrac{-1}{b-a}+\dfrac{1}{b-c}+\dfrac{-1}{c-b}+\dfrac{1}{c-a}\)

\(VT=\dfrac{2}{a-b}+\dfrac{2}{b-c}+\dfrac{2}{c-a}=VP\)