a) \(\frac{3}{4}-1\frac{1}{2}+0,5\div\frac{5}{12}=\frac{3}{4}-\frac{3}{2}+\frac{6}{5}=\frac{9}{20}\)

b) \(\left(-2\right)^2-1\frac{5}{27}\left(-\frac{3}{2}\right)^3=4-\frac{32}{27}.\frac{27}{8}=4-4=0\)

c) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

a/ góc AOB;BOC;COD;AOC;BOD;AOD

b/ góc ABC;BCD;ABD;ACD

Bài 1:

a) 

Bài 1:

a) Phần bù của \(\frac{2^{2020}+9}{2^{2020}+10}\)là \(\frac{1}{2^{2020}+10}\)

Phần bù của \(\frac{2^{2022}+9}{2^{2022}+10}\)là \(\frac{1}{2^{2022}+10}\)

Ta có: \(\frac{1}{2^{2020}+10}>\frac{1}{2^{2022}+10}\)

Do đó: \(\frac{2^{2020}+9}{2^{2020}+10}< \frac{2^{2022}+9}{2^{2022}+10}\)

e) Ta áp dụng kết quả phụ nếu 0<a<b và m>0 thì \(\frac{a}{b}>\frac{a-m}{b-m}\)

\(\frac{2018^{99}-1}{2018^{100}-1}>\frac{2018^{99}-1-2017}{2018^{100}-1-2017}=\frac{2018^{99}-2018}{2018^{100}-2018}=\frac{2018.\left(2018^{98}-1\right)}{2018.\left(2018^{99}-1\right)}=\frac{2018^{98}-1}{2018^{99}-1}\)