Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)

Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)

\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)

\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)

\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)

\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)

Bạn kiểm tra lại đề bài câu 1, câu này chỉ có thể rút gọn đến \(2cot^2x+2cotx+1\) nên biểu thức ko hợp lý

Đồng thời kiểm tra luôn đề câu 2, trong cả 2 căn thức đều xuất hiện \(6sin^2x\) rất không hợp lý, chắc chắn phải có 1 cái là \(6cos^2x\)

Câu 1 đề vẫn có vấn đề:

\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2\left(1+cot^2x\right)cot^2x}{\left(tanx-1\right)\left(tan^2x+1\right)cot^2x}=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^2x}{tanx-1}\)

\(=\dfrac{1+cotx}{1-cotx}-\dfrac{2cot^3x}{1-cotx}=\dfrac{1+cotx-2cot^3x}{1-cotx}\)

\(=\dfrac{\left(1-cotx\right)\left(1+2cotx+2cot^2x\right)}{1-cotx}=1+2cotx+2cot^2x\)

Có thể coi như ko thể rút gọn tiếp

2.

\(\sqrt{\left(1-cos^2x\right)^2+6cos^2x+3cos^4x}+\sqrt{\left(1-sin^2x\right)^2+6sin^2x+3sin^4x}\)

\(=\sqrt{4cos^4x+4cos^2x+1}+\sqrt{4sin^4x+4sin^2x+1}\)

\(=\sqrt{\left(2cos^2x+1\right)^2}+\sqrt{\left(2sin^2x+1\right)^2}\)

\(=2\left(cos^2x+sin^2x\right)+2=4\)

a) \(\left|sinx-cosx\right|+\left|sinx+cosx\right|=2\)

\(\Leftrightarrow\left(sinx-cosx\right)^2+2\left|sinx-cosx\right|\left|sinx+cosx\right|+\left(cosx+sinx\right)^2=4\)

\(\Leftrightarrow2\left(sin^2x+cos^2x\right)+2\left|\left(sinx-cosx\right)\left(sinx+cosx\right)\right|=4\)

\(\Leftrightarrow\left|sin^2x-cos^2x\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=1\\sin^2x-cos^2x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=sin^2x+cos^2x\\sin^2x-cos^2x=-\left(sin^2x+cos^2x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=0\\sin^2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=0\end{matrix}\right.\)\(\Rightarrow cosx.sinx=0\Rightarrow sin2x=0\)

\(\Rightarrow x=\dfrac{k\pi}{2},k\in Z\)

Vậy...

b) ĐK:\(x\ne\dfrac{k\pi}{2};k\in Z\)

Pt \(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cosx}{sinx}=4\left(sinx+\sqrt{3}cosx\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}cosx=0\left(1\right)\\\dfrac{sinx-\sqrt{3}cosx}{sinx.cosx}=4\left(2\right)\end{matrix}\right.\)

Từ \(\left(1\right)\Leftrightarrow tanx=-\sqrt{3}\Leftrightarrow x=-\dfrac{\pi}{3}+k\pi,k\in Z\)

Từ (2)\(\Leftrightarrow sinx-\sqrt{3}cosx=4sinx.cosx\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=2sinx.cosx\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin2x\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy \(\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)

c) ĐK: \(x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\left(k\in Z\right)\)

Pt \(\Leftrightarrow\left(\sqrt{2}sinx-1\right)^2+\left(\sqrt{3}tan2x-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}sinx-1=0\\\sqrt{3}tan2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}sinx=\dfrac{1}{\sqrt{2}}\\tan2x=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)

Vậy pt vô nghiệm

b.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\sqrt{2}\left(sinx+cosx\right)=\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx}\)

\(\Leftrightarrow\sqrt{2}\left(sinx+cosx\right)=\dfrac{1}{sinx.cosx}\)

Đặt \(sinx+cosx=t\Rightarrow\left|t\right|\le\sqrt{2}\)

\(sinx.cosx=\dfrac{t^2-1}{2}\)

Pt trở thành:

\(\sqrt{2}t=\dfrac{2}{t^2-1}\Rightarrow t^3-t-\sqrt{2}=0\)

\(\Leftrightarrow\left(t-\sqrt{2}\right)\left(t^2+\sqrt{2}t+1\right)=0\)

\(\Leftrightarrow t=\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+k2\pi\)

a.

\(\Leftrightarrow sin^22x+cos^22x+\sqrt{3}sin4x+1+cos4x=0\)

\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-2\)

\(\Leftrightarrow\dfrac{1}{2}cos4x+\dfrac{\sqrt{3}}{2}sin4x=-1\)

\(\Leftrightarrow cos\left(4x-\dfrac{\pi}{3}\right)=-1\)

\(\Leftrightarrow4x-\dfrac{\pi}{3}=\pi+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\)

Này là kiến thức lớp 10 mà bạn...