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

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

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

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

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

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

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

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

AC =3cm

Đề bài đâu z :v 

Hỏi z thôi chứ e chưa học lớp 9

\(\frac{x^2}{\sqrt{x+1}+1}=x-4\)

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

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

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

\(\Leftrightarrow2x-4\sqrt{x+1}-4=x^2\)

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

\(\Leftrightarrow2\left(x+1+2.\sqrt{x+1}.1+1\right)=x^2\)

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

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Nếu mk lm sai ở đâu thì cứ ns mk , mk sẽ sửa lại chứ đừng vô cớ k sai nhé , thanks nhìu......