Câu 1: B

Câu 2: C

Câu 3: B

Câu 4: A

Câu 5: C

Câu 6: C

Có : 0 < \(\alpha< \dfrac{\pi}{2}\Rightarrow cos\alpha>0\)  \(\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{4}\right)^2}=\dfrac{\sqrt{7}}{4}\)

\(\Rightarrow sin2\alpha=2sin\alpha.cos\alpha=2.\dfrac{3}{4}.\dfrac{\sqrt{7}}{4}=\dfrac{3\sqrt{7}}{8}\) 

\(cos2\alpha=1-2sin^2\alpha=1-2.\left(\dfrac{3}{4}\right)^2=-\dfrac{1}{8}\)

Ta có : \(A=\dfrac{cot2\alpha}{tan2\alpha+sin2\alpha}=\dfrac{1}{sin2\alpha+\dfrac{sin^22\alpha}{cos2\alpha}}=\dfrac{1}{\dfrac{3\sqrt{7}}{8}+\dfrac{\dfrac{63}{64}}{-\dfrac{1}{8}}}=\dfrac{8}{-63+3\sqrt{7}}\)

Sửa lại : \(A=\dfrac{cot2\alpha}{tan2\alpha+sin2\alpha}=\dfrac{1}{tan^22\alpha+\dfrac{sin^22\alpha}{cos2\alpha}}=\dfrac{1}{\left(-3\sqrt{7}\right)^2+\left(\dfrac{3\sqrt{7}}{8}\right)^2:-\dfrac{1}{8}}=\dfrac{8}{441}\)

Đặt \(A=5\cdot7^{2\left(n+1\right)}+2^{3n}=5\cdot49^{n+1}+8^n=5\left(41+8\right)^{n+1}+8^n\)

Áp dụng công thức nhị thức Newton, ta có:

\(\left(41+8\right)^{n+1}=41^{n+1}+\left(n+1\right)\cdot41^n\cdot8+\dfrac{n\left(n+1\right)}{2}\cdot41^{n-1}\cdot8^2+...+\left(n+1\right)\cdot41\cdot8^n+8^{n+1}\)

Vậy \(A=5\left[41^{n+1}+\left(n+1\right)\cdot41^n\cdot8+..+\left(n+1\right)\cdot41\cdot8^n+8^{n+1}\right]+8^n\)

\(\Rightarrow A=5\left[41^{n+1}\left(n+1\right)\cdot41^n\cdot8+...+\left(n+1\right)\cdot41\cdot8^n\right]+5\cdot8^{n+1}+8^n\)

Đặt \(B=41^{n+1}\left(n+1\right)\cdot41^n\cdot8+...+\left(n+1\right)\cdot41\cdot8^n\)

\(\Rightarrow B⋮41\)

Đặt \(C=5\cdot8^{n+1}+8^n=8^n\left(5\cdot8+1\right)=8^n\cdot41\)

\(\Rightarrow C⋮41\)

Mà \(A=B+C\Rightarrow A⋮41\)

\(\RightarrowĐPCM\)

ý 5 ấy, bạn xem có đúng ko chứ toi cx ko bt đâu :<

a: \(\Leftrightarrow2\sqrt{x^2-3}=3\)

=>căn x^2-3=3/2

=>x^2-3=9/4

=>x^2=9/4+3=21/4

=>\(x=\pm\dfrac{\sqrt{21}}{2}\)

b: =>3 căn 4-x^2=-1(loại)

c: =>x^2-2=3-x^2

=>2x^2=5

=>x^2=5/2

=>\(x=\pm\sqrt{\dfrac{5}{2}}\)