\(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)

\(=\left(\dfrac{a}{c+b}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)

\(=\dfrac{a+c+b}{c+b}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3\)

\(=4034.\dfrac{1}{2}-3=2014\)

Guể?

\(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}=\dfrac{1}{2}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=\dfrac{4034}{2}=2017\)

\(\Rightarrow1+\dfrac{a}{c+b}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}=2017\)

\(\Rightarrow\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=2014\)

2:

\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)

1) Ta có 

\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)

\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}\)

\(C=\dfrac{1}{2022}\)

2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)

\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)

a) \(\dfrac{2}{3}\times\dfrac{1}{4}-\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{2}{12}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=\dfrac{0}{6}=0\)

b) \(\dfrac{8}{5}\times\dfrac{1}{4}-\dfrac{2}{5}\times\dfrac{1}{2}-\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{8}{20}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{2}{10}-\dfrac{1}{10}=\dfrac{4-2-1}{10}=\dfrac{1}{10}\)

CHÚC EM HỌC TỐT NHÉ

a. 1/6 - 1/6 = 0 (hoặc 2/12 - 1/6= 2/12 - 2/12 = 0)

b. 4/5 - 1/2 x ( 2/5 - 1/5 ) = 4/5 - 1/2 x 1/5 

                                        = 4/5 x 2/10

                                        = 4/25

\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)

\(=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)

\(\Rightarrow A< 1.\left(\dfrac{1}{2.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(\Rightarrow A< 1+\left(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Rightarrow A< 1+\left(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}\right)\)

Mà ta thấy \(\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow A< 1+\dfrac{3}{4}=\dfrac{7}{4}\)