Áp dụng định lý Pytago

\(AC=\sqrt{5^2-3^2}=4cm\)

Ta có:

\(cotB=\dfrac{AB}{AC}=\dfrac{3}{4}\)

\(cotC=\dfrac{AC}{AB}=\dfrac{4}{3}\)

\(\Rightarrow P=\dfrac{3}{4}+\dfrac{4}{3}=\dfrac{25}{12}\)

=> \(\widehat{C}=180-90-15=70^o\)

Ta có:

\(sin15=\dfrac{AC}{BC}\Rightarrow BC=\dfrac{AC}{sin15}=\dfrac{4AC}{\sqrt{6}-\sqrt{2}}\)

\(sin75=\dfrac{AB}{BC}\Rightarrow BC=\dfrac{AB}{sin75}=\dfrac{4AB}{\sqrt{6}+\sqrt{2}}\)

\(\Rightarrow BC^2=\dfrac{16.AB.AC}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}=4.AB.AC\)

\(2S=2+2^2+2^3+2^{2018}\)

\(2S-S=2+2^2+2^3+...+2^{2018}-\left(1+2+2^2+...+2^{2017}\right)\)

\(S=2^{2018}-1\)

Áp dụng định lý Pytago

\(AB=\sqrt{25^2-15^2}=20\)

Ta có:

\(sinC=\dfrac{AB}{BC}=\dfrac{20}{25}=\dfrac{4}{5}\)

\(\Rightarrow\widehat{C}\simeq53,1^o\)

\(15-2\left(2x+1\right)=8+3x\)

\(2\left(2x+1\right)=15-\left(8+3x\right)\)

\(2\left(2x+1\right)=7-3x\)

\(4x+2=7-3x\)

\(7x=5\)

\(x=\dfrac{5}{7}\)

\(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+..+\dfrac{1}{97.99}+\dfrac{1}{98.100}-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{1}{97.99}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{99.100}\right)\right]-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{98}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[1-\dfrac{1}{99}+\dfrac{1}{2}-\dfrac{1}{100}\right]-\dfrac{49}{99}\)

\(=\dfrac{1}{2}\left[\dfrac{98}{99}+\dfrac{49}{100}\right]-\dfrac{49}{99}=\dfrac{14651}{19800}-\dfrac{49}{99}=\dfrac{49}{200}\)

\(A=1+2+2^2+2^3+..+2^{62}+2^{63}\)

\(2A=2+2^2+2^3+...+2^{63}+2^{64}\)

\(2A-A=2^{64}-1\)

\(A=2^{64}-1\)

\(A=1+2+2^2+...+2^{63}\)

\(2A=2+2^2+2^3+...+2^{64}\)

\(2A-A=2+2^2+2^3+...+2^{64}-\left(1+2+2^2+...+2^{63}\right)\)

\(\Rightarrow A=2^{64}-1< 2^{64}\)

\(\Rightarrow A< B\)