1.\(sin^2\alpha+cos^2\alpha=\left(\dfrac{AC}{BC}\right)^2+\left(\dfrac{AB}{BC}\right)^2\)

=\(\dfrac{AC^2+AB^2}{BC^2}=\dfrac{BC^2\left(pytago\right)}{BC^2}=1\)

2.ta có \(tan\alpha=\dfrac{AC}{AB}\)

\(\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{AC}{BC}}{\dfrac{AB}{BC}}=\dfrac{AC}{AB}\)

\(\Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)

3.ta có:\(1+tan^2\alpha=1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2\)

=\(\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}\)=\(\dfrac{1}{cos^2\alpha}\)

4.ta có :\(cot\alpha=\dfrac{AB}{AC}\)

\(\dfrac{cos\alpha}{sin\alpha}=\dfrac{\dfrac{AB}{BC}}{\dfrac{AC}{BC}}=\dfrac{AB}{AC}\)

\(\Rightarrow cot\alpha=\dfrac{cos\alpha}{sin\alpha}\)

\(1+cot^2\alpha=1+\left(\dfrac{cos\alpha}{sin\alpha}\right)^2=\dfrac{sin^2\alpha+cos^2\alpha}{sin^2\alpha}\)=\(\dfrac{1}{sin^2a}\)

bạn hỏi quanda ấy 

aor ma panda

Tra trên google hoặc hỏi Quanda đi

\(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\\ A=\sqrt{8-2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\left(4+\sqrt{15}\right)\\ A=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\left(4+\sqrt{15}\right)\\ A=\left(\sqrt{5}-\sqrt{3}\right)^2\left(4+\sqrt{15}\right)\\ A=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\\ A=2\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)=2\left[4^2-\left(\sqrt{15}\right)^2\right]=2\cdot1=2\)

\(A=\sqrt{4-\sqrt{15}}\cdot\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\)

\(=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)

\(=2>\sqrt{3}\)

xét vế trái ta có

\(A=\sqrt{4-\sqrt{15}}.\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\)

A=\(\sqrt{4^2-\sqrt{15}^2}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\)

A=\(\sqrt{4+\sqrt{15}}.\sqrt{\sqrt{10}-\sqrt{6}}.\sqrt{10-\sqrt{6}}\)

A=\(\sqrt{\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)}.\sqrt{\sqrt{10}-\sqrt{6}}\)

A=\(\sqrt{4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}}.\sqrt{\sqrt{10}-\sqrt{6}}\)

A=\(\sqrt{4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}}.\sqrt{\sqrt{10}-\sqrt{6}}\)

A=\(\sqrt{\sqrt{10}+\sqrt{6}}.\sqrt{\sqrt{10}-\sqrt{6}}\)

A=\(\sqrt{\sqrt{10}^2-\sqrt{6}^2}=\sqrt{4}\)

mà:\(\sqrt{4}>\sqrt{3}\) nên A\(>\sqrt{3}\)

Đặt\(\begin{cases} x+y=S \\ xy=P \end{cases}\)

Ta có:\(\begin{cases} S-2P=0 \\ S-P^2=\sqrt{(P-1)^2+1} \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S=2P \\ 2P-P^2=\sqrt{(P-1)^2+1} \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S=2P \\ (2P-P^2)^2=(P-1)^2+1 \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S=2P \\ 4P^2-4P^3+P^4=P^2-2P+2 \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S=2P \\ P^4-4P^3+3P^2+2P-2=0 \end{cases}\)

\(\Leftrightarrow\)\(\begin{cases} S =2+2\sqrt{3}\\ P=1+\sqrt{3} \end{cases}\)(1)hoặc\(\begin{cases} S=2 \\ P=1 \end{cases}\)(2)hoặc\(\begin{cases} S=2-2 \sqrt{3}\\ P=1-\sqrt{3} \end{cases}\)(3)

Còn lại là thay vào biểu thức x²-Sx+P=0 thôi