Bài 1:
a)
ta có: \(\dfrac{50}{100}=\dfrac{1}{2};\dfrac{-\dfrac{4}{13}}{-\dfrac{8}{13}}=\dfrac{1}{2};\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{1}{2};\dfrac{-\dfrac{2}{17}}{-\dfrac{4}{17}}=\dfrac{1}{2}\)
\(\dfrac{50}{100}=\dfrac{\dfrac{4}{13}}{\dfrac{8}{13}}=\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{\dfrac{2}{17}}{\dfrac{4}{17}}=\dfrac{50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}=\dfrac{1}{2}\)
vậy \(A=\dfrac{1}{2}\)
b)
\(B=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\\ B=\dfrac{1}{19}-\dfrac{1}{19}+\dfrac{2}{29}-\dfrac{2}{29}+\dfrac{3}{39}-...-\dfrac{199}{1999}+\dfrac{200}{2009}\\ B=\dfrac{200}{2009}\)
Bài 2:
\(\dfrac{a}{b}=\dfrac{b}{3c}=\dfrac{c}{9a}=\dfrac{b+c}{3c+9a}\)
suy ra: \(b=\dfrac{3c\left(b+c\right)}{3c+9a}=\dfrac{3cb+3c^2}{3c+9a}=\dfrac{bc+c^2}{c+3a}\)
\(c=\dfrac{9a\left(b+c\right)}{3c+9a}=\dfrac{9ab+9ac}{3c+9a}=\dfrac{3ab+3ac}{c+3a}\)
giả sử b=c là đúng thì :\(\dfrac{bc+c^2}{c+3a}=\dfrac{3ab+3ac}{c+3a}\)
hay \(bc+c^2=3ab+3ac\\ \Leftrightarrow c^2+bc-3ab-3ac=0\)
\(\Leftrightarrow\left(b+c\right)\left(c-3a\right)=0\Rightarrow c-3a=0\Rightarrow c=3a\)
b) \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}+\dfrac{2}{2014.2016}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2016}\right)=\dfrac{2015}{4032}< 1\)
mà \(1< \dfrac{4}{3}\) nên \(\dfrac{2015}{4032}< \dfrac{4}{3}\)
hay \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}< \dfrac{4}{3}\)
bài 3:
a)\(\left(x-y\right)\left(x+y\right)=x^2-y^2-xy+xy=x^2-y^2\) (đpcm)
b) áp dụng BĐT tam giác, ta có:
\(a+b>c\Rightarrow a+b-c>0\\ b+c>a\Rightarrow b+c-a< 0\\ a+c>b\Rightarrow a-b+c>0\)
suy ra: \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< 0\: \: \: \: \: \: \)
đồng thời \(abc>0\) với mọi a, b, c dương.
nên \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< abc\)
ko tìm dc dấu bằng xảy ra.
