10.

\(\dfrac{sin3x-cos3x}{sinx+cosx}=\dfrac{3sinx-4sin^3x-\left(4cos^3x-3cosx\right)}{sinx+cosx}\)

\(=\dfrac{3\left(sinx+cosx\right)-4\left(sin^3x+cos^3x\right)}{sinx+cosx}\)

\(=\dfrac{3\left(sinx+cosx\right)-4\left(sinx+cosx\right)\left(sin^2x+cos^2x-sinx.cosx\right)}{sinx+cosx}\)

\(=\dfrac{3\left(sinx+cosx\right)-4\left(sinx+cosx\right)\left(1-sinx.cosx\right)}{sinx+cosx}\)

\(=\dfrac{\left(sinx+cosx\right)\left(3-4+4sinx.cosx\right)}{sinx+cosx}\)

\(=-1+4sinx.cosx\)

\(=2sin2x-1\)

11.

\(tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\dfrac{1+cos\left(\dfrac{\pi}{2}+x\right)}{sin\left(\dfrac{\pi}{2}+x\right)}=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{1+sin\left(-x\right)}{cos\left(-x\right)}\)

\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{1-sinx}{cosx}=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\dfrac{sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}-2sin\dfrac{x}{2}cos\dfrac{x}{2}}{cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}}\)

\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)^2}{\left(cos\dfrac{x}{2}-sin\dfrac{x}{2}\right)\left(cos\dfrac{x}{2}+sin\dfrac{x}{2}\right)}\)

\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{cos\dfrac{x}{2}-sin\dfrac{x}{2}}{cos\dfrac{x}{2}+sin\dfrac{x}{2}}\)

\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).\dfrac{cos\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)}{sin\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)}\)

\(=tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).cot\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\)

\(=1\)

12.

\(cot2x+\dfrac{1}{sin2x}+tanx=\dfrac{cos2x}{sin2x}+\dfrac{1}{sin2x}+tanx\)

\(=\dfrac{cos2x+1}{sin2x}+\dfrac{sinx}{cosx}=\dfrac{2cos^2x-1+1}{2sinx.cosx}+\dfrac{sinx}{cosx}\)

\(=\dfrac{2cos^2x}{2sinx.cosx}+\dfrac{sinx}{cosx}=\dfrac{cosx}{sinx}+\dfrac{sinx}{cosx}\)

\(=\dfrac{sin^2x+cos^2x}{sinx.cosx}=\dfrac{1}{sinx.cosx}\)

\(=\dfrac{2}{2sinx.cosx}=\dfrac{2}{sin2x}\)

17.

Câu 17 này đề bài sai (ở độ dài AB, nếu ko nhìn lầm thì AB=41 là 1 con số phi lý)

Cách tính như sau:

Qua \(C_1\) kẻ các đường thẳng song song AC và BC, cắt \(AA_1\) và \(BB_1\) kéo dài tại D và E

\(\Rightarrow ABC.DEC_1\) là lăng trụ đứng có thể tích V

\(V=CC_1.S_{ABC}=6.\dfrac{AB^2\sqrt{3}}{4}=\dfrac{3AB^2\sqrt{3}}{2}\)

Gọi thể tích khối đa diện cần tính là \(V_1\)

\(\Rightarrow\dfrac{V_1}{V}=\dfrac{1}{3}\left(\dfrac{AA_1}{AD}+\dfrac{BB_1}{BE}+\dfrac{CC_1}{CC_1}\right)=\dfrac{5}{6}\)

\(\Rightarrow V_1=\dfrac{5}{6}V=...\)

Hình vẽ câu 17:

19.

Gọi chiều cao hình hộp là y và cạnh đáy là x

\(\Rightarrow x^2y=6\Rightarrow y=\dfrac{6}{x^2}\)

\(S_đ=x^2\) ; \(S_{xq}=4xy=4x.\dfrac{6}{x^2}=\dfrac{24}{x}\)

\(\Rightarrow S_đ+S_{xq}=x^2+\dfrac{24}{x}=x^2+\dfrac{12}{x}+\dfrac{12}{x}\ge3\sqrt[3]{\dfrac{144x^2}{x^2}}=3\sqrt[3]{144}\)

Dấu "=" xảy ra khi \(x^2=\dfrac{12}{x}\Rightarrow x=\sqrt[3]{12}\)