a) Để A có nghĩa :

\(\Rightarrow\sqrt{2x+3-x^2\: }\Leftrightarrow2+\sqrt{2x+3-x^2}\ge2\forall x\) 

\(\Rightarrow\sqrt{-\left(x-1\right)^2+4}\ge0\) 

\(\Leftrightarrow-\left(x-1\right)^2\ge-4\) 

\(\Leftrightarrow\left(x-1\right)^2\le4\) 

\(\Rightarrow3\ge x\ge-1\) 

Vậy.....

5.What are you doing at the moment?

...........I am playing football at the moment.........................

6.Does your sister often practise English?

...........Yes,she does....................

7.Who read you diary two days ago?

................Nam read your diary two days ago.....................

8.Where did you see this advertisement?

.........I saw this advertisement on the streets........

Theo mình nghĩ thì đề bài là trả lời câu hỏi.Ko phải thì nhắn tin cho mình để mình làm lại nhé!

\(=\left(\sqrt{3}-1\right).\sqrt{2}.\sqrt[]{2+\sqrt{3}}\)

\(=\left(\sqrt{3}-1\right).\sqrt[]{2.\left(2+\sqrt{3}\right)}\)

\(=\left(\sqrt{3}-1\right).\sqrt{4+2\sqrt{3}}\)

\(=\left(\sqrt{3}-1\right).\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}\)

\(=\left(\sqrt{3}-1\right).\sqrt{3+2\sqrt{3}+1}\)

\(=\left(\sqrt{3}-1\right).\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left(\sqrt{3}-1\right).|\sqrt{3}+1|\)

\(=\left(\sqrt{3}-1\right).\left(\sqrt{3}+1\right)\)

\(=\left(\sqrt{3}\right)^2-1\)

\(=3-1\)

\(=2\)

Câu hỏi của Le Ngan - Toán lớp 9 - Học toán với OnlineMath

vì 2016 \(⋮\)4 nên đặt a²⁰¹⁶ = a^4k sau đó làm tương tự 

\(x^2-16+2\left(x+4\right)\)

\(=\left(x+4\right)\left(x-4\right)+2\left(x+4\right)\)

\(=\left(x+4\right)\left(x-4+2\right)\)

\(=\left(x+4\right)\left(x-2\right)\)

\(x^2-16+2\left(x+4\right)=x^2+2x-8=x^2-2x+4x-8\)

\(=x\left(x-2\right)+4\left(x-2\right)=\left(x+4\right)\left(x-2\right)\)

Áp dụng hệ thức lượng trong tam giác vuông ta có

$\begin{array}{c}TheoPytago:\\ B{C}^2=A{B}^2+A{C}^2={15}^2+{20}^2=625\\ \Rightarrow BC=25\left(cm\right)\\ Trong:\Delta BCD\perp C;CA\perp BD\\ \Rightarrow B{C}^2=BA.BD\\ \Rightarrow BD=\frac{625}{15}=\frac{125}{3}\left(cm\right)\\ \Rightarrow AD=BD-AB=\frac{125}{3}-15=\frac{80}{3}\left(cm\right)\\ \Rightarrow CD=\sqrt{DA.DB}=\sqrt{\frac{80}{3}.\frac{125}{3}}=\frac{100}{3}\left(cm\right)\end{array}$

a) \(\tan^2\alpha+1=\frac{\sin^2\alpha}{\cos^2\alpha}+1=\frac{\sin^2\alpha+\cos^2\alpha}{\cos^2\alpha}=\frac{1}{\cos^2\alpha}\)

b) \(\cot^2\alpha+1=\frac{\cos^2\alpha}{\sin^2\alpha}+1=\frac{\cos^2\alpha+\sin^2\alpha}{\sin^2\alpha}=\frac{1}{\sin^2\alpha}\)

c) \(\cos^4\alpha-\sin^4\alpha=\left(\cos^2\alpha+\sin^2\alpha\right)\left(\cos^2\alpha-\sin^2\alpha\right)=\cos^2\alpha-\sin^2\alpha\)

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