5 7 − 7 5 − 1 2 − − 2 7 − 1 10 = − 24 35 − 1 2 − − 27 70 = − 24 35 − 31 35 = − 11 7

a)  − 5 7 + − 6 8 = − 5 7 + − 3 4 =   − 20 28 + − 21 28 = − 41 28   = − 123 84

− 1 3 = − 28 84

Vậy  − 5 7 + − 6 8     < − 28 84

b)  14 20 + 4 − 5     = 7 10 + − 4 5 = 7 10 + − 8 10 = − 1 10

Vậy  14 20 + 4 − 5     = − 1 10

Số đo góc ngoài tại đỉnh B là:

\(180^0-360^0+57^0+110^0+75^0=62^0\)

1

\(=\dfrac{1}{110}\times\dfrac{1}{1}\times\dfrac{1}{1}\times\dfrac{114}{1}\times\dfrac{55}{57}\\ =\dfrac{114\times55}{110\times57}\\ =\dfrac{2\times1}{2\times1}=\dfrac{2}{2}=1\)

\(\dfrac{111}{110}\times\dfrac{112}{111}\times\dfrac{113}{112}\times\dfrac{114}{113}\times\dfrac{55}{57}=\dfrac{111\times112\times113\times114}{110\times111\times112\times113}\times\dfrac{55}{57}\)

                                                   \(=\dfrac{114}{110}\times\dfrac{55}{57}=\dfrac{57}{55}\times\dfrac{55}{57}=1\)

Số đo góc ngoài tại đỉnh B là:

180-110=70(độ)

A - B = \(\left(1+\frac{1}{2}+1+\frac{1}{12}+1+\frac{1}{30}+1+\frac{1}{56}+1+\frac{1}{90}\right)-\left(1-\frac{1}{6}+1-\frac{1}{20}+1-\frac{1}{42}+1-\frac{1}{72}+1-\frac{1}{110}\right)\)= \(\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)-\left(5-\frac{1}{6}-\frac{1}{20}-\frac{1}{42}-\frac{1}{72}-\frac{1}{110}\right)\)\

= \(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}-5+\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\)

= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\)

Xét tứ giác ABCD có :

∠A+∠B+∠C+∠D = 360^o

=>57^o+∠B+110^o+75^o = 360^o

=>∠B = 118^o

Số đo góc ngoài của đỉnh B là : 180^o-∠B = 180^o-118^o = 62^o

(Góc trong tứ giác ABCD là \(\widehat{B}\), góc ngoài tại đỉnh B là \(\widehat{B_1}\))

Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\) (t/c tứ giác)

\(\Rightarrow\widehat{B}=360^o-\widehat{A}-\widehat{C}-\widehat{D}\)

\(\Rightarrow\widehat{B}=360^o-57^o-110^o-75^o\)

\(\Rightarrow\widehat{B}=118^o\)

mà \(\widehat{B}+\widehat{B_1}=180^o\) (kề bù)

\(\Rightarrow\widehat{B_1}=62^o\)

\(A=\frac{3}{2}-\frac{5}{6}+\frac{13}{12}-\frac{19}{20}+\frac{31}{30}-\frac{41}{42}+\frac{57}{56}-\frac{71}{72}+\frac{91}{90}-\frac{109}{110}\)

\(\Rightarrow A=\left(1+\frac{1}{2}\right)-\left(1-\frac{1}{6}\right)+\cdot\cdot\cdot+\left(1+\frac{1}{90}\right)-\left(1-\frac{1}{110}\right)\)

\(\Rightarrow A=1+\frac{1}{2}-1+\frac{1}{6}+\cdot\cdot\cdot+1+\frac{1}{90}-1+\frac{1}{110}\)

\(\Rightarrow A=\left[\left(1-1\right)+\frac{1}{2}+\frac{1}{6}\right]+\cdot\cdot\cdot+\left[\left(1-1\right)+\frac{1}{90}+\frac{1}{110}\right]\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{90}+\frac{1}{110}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow A=1-\frac{1}{11}\)

\(\Rightarrow A=\frac{10}{11}\)