\(E=\dfrac{\left(cosx-siny\right)\left(cosx+siny\right)}{sin^2x\cdot sin^2y}-\dfrac{cos^2x}{sin^2x}\cdot\dfrac{cos^2y}{sin^2y}\)

\(=\dfrac{cos^2x\left(1-cos^2y\right)-sin^2y}{sin^2x\cdot sin^2y}\)

\(=\dfrac{sin^2y\left(cos^2x-1\right)}{sin^2x\cdot sin^2y}=-1\)

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

\(cos^4x+sin^2x.cos^2x+sin^2x\)

\(=cos^2x.cos^2x+sin^2x.cos^2x+sin^2x\)

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

\(=cos^2x.1+sin^2x\)

\(=cos^2x+sin^2x\)

\(=1\)

a: \(A=2\left(\sin^6x+cos^6x\right)-3\cdot\left(\sin^4x+cos^4x\right)\)

\(=2\cdot\left\lbrack\left(\sin^2x+cos^2x\right)^3-3\cdot\sin^2x\cdot cos^2x\cdot\left(\sin^2x+cos^2x\right)\right\rbrack-3\cdot\left\lbrack\left(sin^2x+cos^2x\right)^2-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

\(=2\left\lbrack1-3\cdot sin^2x\cdot cos^2x\right\rbrack-3\cdot\left\lbrack1-2\cdot\sin^2x\cdot cos^2x\right\rbrack\)

\(=2-6\cdot\sin^2x\cdot cos^2x-3+6\cdot\sin^2x\cdot cos^2x\)

=2-3

=-1

c: \(C=\frac{\sin^2x}{1+\cot x}+\frac{cos^2x}{1+\tan x}+\sin x\cdot cosx\)

\(=\frac{\sin^2x}{1+\frac{cosx}{\sin x}}+\frac{cos^2x}{1+\frac{\sin x}{cosx}}+\sin x\cdot cosx=\sin^2x:\frac{\sin x+cosx}{\sin x}+cos^2x:\frac{\sin x+cosx}{cosx}+\sin x\cdot cosx\)

\(=\frac{\sin^3x+cos^3x}{\sin x+cosx}+\sin x\cdot cosx\)

\(=\frac{\left(\sin x+cosx\right)\left(\sin^2x-\sin x\cdot cosx+cos^2x\right)}{\sin x+cosx}+\sin x\cdot cosx\)

\(=\sin^2x-\sin x\cdot cosx+cos^2x+\sin x\cdot cosx\)

\(=\sin^2x+cos^2x=1\)

d: \(D=\frac{\cot^2x-cos^2x}{cot^2x}+\frac{\sin x\cdot cosx}{\cot x}\)

\(=\left(\frac{cos^2x}{\sin^2x}-cos^2x\right):\frac{cos^2x}{sin^2x}+\frac{\sin x\cdot cosx}{\frac{cosx}{\sin x}}\)

\(=cos^2x\left(\frac{1}{\sin^2x}-1\right)\cdot\frac{\sin^2x}{cos^2x}+\frac{\sin x\cdot cosx\cdot\sin x}{cosx}\)

\(=\frac{1-\sin^2x}{\sin^2x}\cdot\sin^2x+\sin^2x=1-\sin^2x+\sin^2x=1\)

\(1.\)

ĐKXĐ : \(x\ge4\)

\(pt\Leftrightarrow\sqrt{x+4}+\sqrt{x-4}=2\sqrt{x^2-16}+2x-12\)

\(\Leftrightarrow\sqrt{x+4}+\sqrt{x-4}=x+4+2\sqrt{\left(x+4\right)\left(x-4\right)}+x-4-12\)

\(\Leftrightarrow\sqrt{x+4}+\sqrt{x-4}=\left(\sqrt{x+4}+\sqrt{x-4}\right)^2-12\) \(\left(1\right)\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=y\) \(\left(y>0\right)\)

\(pt\left(1\right)\Leftrightarrow y=y^2-12\)

\(y^2-y-12=0\)

\(\Leftrightarrow y^2-4y+3y-12=0\)

\(\Leftrightarrow\left(y-4\right)\left(y+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=4\\y=-3\left(\text{loại}\right)\end{matrix}\right.\)

\(y=4\Leftrightarrow\sqrt{x+4}+\sqrt{x-4}=4\)

\(\Leftrightarrow2x+2\sqrt{x^2-16}=16\)

\(\Leftrightarrow\sqrt{x^2-16}=8-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}8-x\ge0\\x^2-16x=x^2-16x+64\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le8\\0x=64\left(\text{vô nghiệm}\right)\end{matrix}\right.\)

Vậy phương trình vô nghiệm

1/ ĐKXĐ: \(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=a>0\)

\(\Rightarrow a^2=2x+2\sqrt{x^2-16}\Rightarrow x+\sqrt{x^2-16}=\frac{a^2}{2}\)

Phương trình trở thành:

\(a=2\left(\frac{a^2}{2}-6\right)\Leftrightarrow a^2-a-6=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+4}+\sqrt{x-4}=3\Rightarrow2x+2\sqrt{x^2-16}=9\)

\(\Rightarrow2\sqrt{x^2-16}=9-2x\) (\(x\le\frac{9}{2}\))

\(\Rightarrow4\left(x^2-16\right)=\left(9-2x\right)^2\)

Phương trình bậc 2 rồi đó, bạn tự giải

2/ Cho T rồi bắt làm gì bây giờ bạn ơi?

3/ Chứng minh cái gì bạn ơi?

4/ Không giải được bạn ơi, pt này chỉ giải được khi x; y là số nguyên tố, không phải số nguyên, mình gặp vài chục lần rồi nên vẫn nhớ :(

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

\(+\left(sin^2x+cos^2x\right)^2-2sin^2x\cdot cos^2x+5\cdot sin^2x\cdot cos^2x\)

\(=sin^4x+cos^4x-sin^2x\cdot cos^2x+1-2\cdot sin^2x\cdot cos^2x+5\cdot sin^2x\cdot cos^2x\)

\(=1-2\cdot sin^2x\cdot cos^2x-sin^2x\cdot cos^2x+1-2\cdot sin^2x\cdot cos^2x+5\cdot sin^2x\cdot cos^2x\)

\(=2\)

Cảm ơn bạn !!!