Dễ thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\)  ; \(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\)

\(\Rightarrow A=\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}=B\)

Sao dài quá z?

haha

bài nó như v

Câu đăng như này thì khoảng 3 - 5 câu 1 lượt thôi nha :)) 

Bài 1.

\(a,\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)

\(b,\dfrac{-2}{5}+\dfrac{7}{21}=\dfrac{-42}{105}+\dfrac{35}{105}=\dfrac{-77}{105}=\dfrac{-11}{15}\)

\(c,\dfrac{3}{8}+\dfrac{-5}{6}=\dfrac{18}{48}+\dfrac{-40}{48}=\dfrac{-22}{48}=\dfrac{-11}{24}\)

\(d,\dfrac{-3}{4}+\dfrac{2}{5}=\dfrac{-15}{20}+\dfrac{8}{20}=\dfrac{-7}{20}\)

\(e,\dfrac{1}{6}+\dfrac{-3}{2}=\dfrac{2}{12}+\dfrac{-18}{12}=\dfrac{16}{12}=\dfrac{4}{3}\)

\(f,\dfrac{-2}{5}+\dfrac{-4}{3}=\dfrac{-6}{15}+\dfrac{-20}{15}=\dfrac{-26}{15}\)

\(g,\dfrac{1}{8}+\dfrac{-3}{4}=\dfrac{4}{32}+\dfrac{-24}{32}=\dfrac{20}{32}=\dfrac{5}{8}\)

\(h,\dfrac{-3}{4}+\dfrac{3}{7}=\dfrac{-21}{28}+\dfrac{12}{28}=\dfrac{-9}{28}\)

\(i,\dfrac{-3}{4}+\dfrac{-4}{5}=\dfrac{-15}{20}+\dfrac{-16}{20}=\dfrac{-31}{20}\)

\(k,\dfrac{-5}{25}+\dfrac{-7}{14}=\dfrac{-70}{350}+\dfrac{-175}{350}=\dfrac{-245}{350}=\dfrac{-7}{10}\)

\(l,\dfrac{6}{21}+\dfrac{-3}{15}=\dfrac{90}{315}+\dfrac{-63}{315}=\dfrac{27}{315}=\dfrac{3}{35}\)

\(m,\dfrac{3}{4}-\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{24}{32}-\dfrac{20}{32}+\dfrac{1}{2}=\dfrac{4}{32}+\dfrac{1}{2}=\dfrac{1}{8}+\dfrac{1}{2}=\dfrac{2}{16}+\dfrac{8}{16}=\dfrac{10}{16}=\dfrac{5}{8}\)

\(n,\dfrac{-3}{12}-\dfrac{1}{-4}+\dfrac{-2}{6}=\dfrac{-1}{4}-\dfrac{1}{-4}+\dfrac{-1}{3}=0+\dfrac{-1}{3}=\dfrac{-1}{3}\)

\(p,\dfrac{1}{3}+\dfrac{-3}{4}-\dfrac{5}{12}=\dfrac{4}{12}+\dfrac{-9}{12}-\dfrac{5}{12}=\dfrac{-5}{12}-\dfrac{5}{12}=\dfrac{-10}{12}=\dfrac{-5}{6}\)

Bài 2:

\(a,\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(x=\dfrac{5}{6}-\dfrac{3}{5}\)

\(x=\dfrac{7}{30}\)

\(b,x-\dfrac{1}{4}=\dfrac{-5}{8}\)

\(x=\dfrac{-5}{8}+\dfrac{1}{4}\)

\(x=\dfrac{-3}{8}\)

\(c,x-\dfrac{2}{3}=\dfrac{7}{12}\)

\(x=\dfrac{7}{12}+\dfrac{2}{3}\)

\(x=\dfrac{5}{4}\)

\(d,x+\dfrac{-7}{15}=-1\dfrac{1}{20}\)

\(x=-1\dfrac{1}{20}-\dfrac{-7}{15}\)

\(x=\dfrac{-7}{12}\)

\(e,x-\dfrac{1}{2}=\dfrac{-3}{4}\)

\(x=\dfrac{-3}{4}+\dfrac{1}{2}\)

\(x=\dfrac{-1}{4}\)

\(f,x-\dfrac{3}{4}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}+\dfrac{3}{4}\)

\(x=\dfrac{5}{4}\)

\(g,x-\dfrac{1}{4}=\dfrac{5}{8}\cdot\dfrac{2}{3}\)

\(x-\dfrac{1}{4}=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}+\dfrac{1}{4}\)

\(x=\dfrac{2}{3}\)

\(h,\dfrac{4}{5}-x=\dfrac{-8}{35}\)

\(x=\dfrac{4}{5}-\dfrac{-8}{35}\)

\(x=\dfrac{36}{35}\)

#YVA

b:

Ta có: MN\(\perp\)AC

AB\(\perp\)AC

Do đó: MN//AB

Xét ΔACB có

M là trung điểm của BC

MN//AB

Do đó: N là trung điểm của AC

=>\(\dfrac{AN}{AC}=\dfrac{1}{2}\)

c:

Ta có: ΔABC vuông tại A

mà AM là đường trung tuyến

nên MA=MB=MC

=>MA=MB

=>ΔMAB cân tại M

Ta có: \(\widehat{DAB}+\widehat{MAB}=\widehat{MAD}=90^0\)

\(\widehat{HAB}+\widehat{HBA}=90^0\)(ΔHAB vuông tại H)

mà \(\widehat{MAB}=\widehat{HBA}\)(ΔMAB cân tại M)

nên \(\widehat{DAB}=\widehat{HAB}\)

=>AB là tia phân giác của góc DAH

It is believed that he won the prize in the contest yesterday.

He is believed to have won the prize in the contest yesterday.