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

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

Câu 1:

Để giá trị của phân thức được xác định:

\(\Rightarrow2-x\ne0\\ \Rightarrow x\ne2\)

\(\RightarrowĐáp.án.B\)

Câu 2: C

Câu 3:C

Đầu tiên ta thấy : \(\dfrac{11}{6}>0\) vì đó là số hữu tỉ dương.

\(\dfrac{8}{-9}=\dfrac{-8}{9}< 0\) vì đó là số hữu tỉ âm.

Do đó : \(\dfrac{11}{6}>\dfrac{8}{-9}\)

\(\dfrac{11}{6}>0\)

\(\dfrac{8}{-9}< 0\)

\(\Rightarrow\dfrac{11}{6}>\dfrac{8}{-9}\)

Vì \(\dfrac{11}{6}>1>\dfrac{-8}{9}\) nên \(\dfrac{11}{6}>\dfrac{8}{-9}\)

   Ta có             \(2^{13}.5^8.10\\ \\ =2^8.2^5.5^8.10\\ \\ =\left(2^8.5^8\right).2^5.10\\ \\ =10^8.10.32\\ \\ =10^9.32\\ \\ =32000000000\)⇒  \(2^{13}.5^8.10\) có 11 chữ số

\(a=2^{13}.5^8.10\)

\(a=2^8.2^5.5^8.10\)

\(a=\left(2^8.5^8\right).2^5.10\)

\(a=\left(2.5\right)^8.32.10\)

\(a=10^8.32.10\)

\(a=1000...0\left(8cs0\right).10.32\)

\(a=1000...0\left(9cs0\right).32\)

\(a=32000...0\left(9cs0\right)\)

\(\Rightarrow a\) có 11 chữ số

a có 11cs

\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)

\(=21-14\sqrt{2}+14\sqrt{2}\)

=21

a) \(\dfrac{35}{101}=\dfrac{105}{303}< \dfrac{189}{303}\Rightarrow\dfrac{35}{101}< \dfrac{189}{303}\)

b) \(\dfrac{11}{13}< \dfrac{11+2}{13+2}=\dfrac{13}{15}< \dfrac{14}{15}\Rightarrow\dfrac{11}{-13}>\dfrac{-14}{15}\)

c) \(-\dfrac{32}{19}< 0< \dfrac{23}{32}\Rightarrow-\dfrac{32}{19}< \dfrac{23}{32}\)

d) \(1,561< 1,5661\Rightarrow-1,561>-1,5661\)

e) \(0,1=\dfrac{1}{10}=\dfrac{40}{400}< \dfrac{40+56}{400+56}=\dfrac{96}{456}< \dfrac{176}{456}\Rightarrow0,1< \dfrac{176}{456}\)

g) \(0,3=\dfrac{3}{10}=\dfrac{9}{30}< \dfrac{9+8}{30+8}=\dfrac{17}{38}< \dfrac{19}{38}\Rightarrow0,3< \dfrac{19}{38}\Rightarrow-0,3>\dfrac{-19}{38}\)

-25/20 < 20/25

15/21 > 21/49

-19/49 > -23/47