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

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

Bài 1:

a: \(\frac35+\frac{5}{13}+\frac45-\frac{18}{13}\)

\(=\left(\frac35+\frac45\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)

\(=\frac75-1=\frac25\)

b: \(\left(\frac37-\frac23\right)^2+\frac{21}{9}=\left(\frac{9}{21}-\frac{14}{21}\right)^2+\frac{21}{9}\)

\(=\left(-\frac{5}{21}\right)^2+\frac73=\frac{25}{441}+\frac73=\frac{1054}{441}\)

c: \(9\cdot\left(-\frac13\right)^3-2\cdot\left(-\frac32\right)^2-\left(-2012\right)^0\)

\(=9\cdot\frac{-1}{27}-2\cdot\frac94-1\)

\(=-\frac13-\frac92-1=-\frac43-\frac92=\frac{-8-27}{6}=-\frac{35}{6}\)

d: \(\left|-\frac43\right|:\left(-2\right)^3-\sqrt{\frac{9}{64}}\)

\(=\frac43:\left(-8\right)-\frac38\)

\(=\frac{-4}{3\cdot8}-\frac38=\frac{-1}{6}-\frac38=\frac{-4}{24}-\frac{9}{24}=-\frac{13}{24}\)

Bài 2:

a: \(x^2=81\)

=>\(\left[\begin{array}{l}x=9\\ x=-9\end{array}\right.\)

b: \(\left(x-1\right)^2=\frac{9}{16}\)

=>\(\left[\begin{array}{l}x-1=\frac34\\ x-1=-\frac34\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac34+1=\frac74\\ x=-\frac34+1=\frac14\end{array}\right.\)

c: ĐKXĐ: x>=0

\(x-2\sqrt{x}=0\)

=>\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}=2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=4\left(nhận\right)\end{array}\right.\)

d: ĐKXĐ:x>=0

\(x=\sqrt{x}\)

=>\(x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=1\left(nhận\right)\end{array}\right.\)

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.