2\(\frac{ }{ }\)
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Tự chứng minh từng cái này rồi suy ra cái đó nhé b.
Ta có: \(sin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}-sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=sin^2\frac{A}{2}\)
Tương tự ta suy ra:
\(sin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}+cos\frac{A}{2}sin\frac{B}{2}cos\frac{C}{2}+cos\frac{A}{2}cos\frac{B}{2}sin\frac{C}{2}=sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}+3sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}\left(1\right)\)
Tiếp theo chứng minh:
\(2sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=\frac{cosA+cosB+cosC-1}{2}\left(2\right)\)
\(sin^2\frac{A}{2}+sin^2\frac{B}{2}+sin^2\frac{C}{2}=\frac{3}{2}-\frac{cosA+cosB+cosC}{2}\left(3\right)\)
\(tan\frac{A}{2}tan\frac{B}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}=1\left(4\right)\)
Từ (1), (2), (3), (4) suy được điều phải chứng minh

\(=\frac{-1\left(\frac{2}{3}-\frac{3}{4}+2\right)}{\frac{2}{3}-\frac{3}{4}+2}-\frac{-1\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=-1-\left(-1\right)\)
\(=-1+1\)
\(=0\)
\(=\frac{-\left(\frac{2}{3}+\frac{3}{4}-2\right)}{\frac{2}{3}+\frac{3}{4}-2}-\frac{-\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=\left(-1\right)-\left(-1\right)\)
\(=0\)

\(1+\frac{1+\frac{1+\frac{3}{2}}{2}}{2}=1+\frac{1+\frac{\frac{5}{2}}{2}}{2}=1+\frac{1+\frac{5}{4}}{2}=1+\frac{\frac{9}{4}}{2}=1+\frac{9}{8}=\frac{17}{8}\)
\(1+\frac{2}{1+\frac{2}{1+\frac{2}{3}}}=1+\frac{2}{1+\frac{2}{\frac{5}{3}}}=1+\frac{2}{1+\frac{6}{5}}=1+\frac{2}{\frac{11}{5}}=1+\frac{10}{11}=\frac{21}{11}\)
\(1+\frac{1+\frac{1+\frac{2}{3}}{3}}{3}=1+\frac{1+\frac{\frac{5}{3}}{3}}{3}=1+\frac{1+\frac{5}{9}}{3}=1+\frac{\frac{14}{9}}{3}=1+\frac{14}{27}=\frac{41}{27}\)
\(\frac{3}{\frac{3}{\frac{3}{\frac{3}{2}+1}+1}+1}+1=1+\frac{3}{\frac{3}{\frac{3}{\frac{5}{2}}+1}+1}=1+\frac{3}{\frac{3}{\frac{6}{5}+1}+1}=1+\frac{3}{\frac{15}{11}+1}=\frac{59}{26}\)
suy ra
\(\frac{\frac{17}{18}}{\frac{21}{11}}-x=\frac{187}{378}-x=\frac{\frac{41}{27}}{\frac{59}{26}}=\frac{1066}{1593}\Rightarrow x=-\frac{1297}{7434}\)

giải:
ta có :
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}\)
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}.\frac{2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{3\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}=\frac{2}{3}\)
đề lỗi rồi bạn