Chủ đề:
Chương I - Căn bậc hai. Căn bậc baCâu hỏi:
Cho tana=\(\dfrac{1}{3}\)Tính\(\dfrac{cosa-sina}{cosa+sina}\)
Chứng minh rằng:\(\dfrac{1-tana}{1+tana}=\dfrac{cosa-sina}{cosa+sina}\)
\(\sqrt{28-6\sqrt{3}}\)
\(\sqrt{6-\sqrt{20}}\)
\(\sqrt{2x+3+2\sqrt{\left(x+1\right)\cdot\left(x+2\right)}}\)
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
\(\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)
A=(\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\))/(\(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\))