a: Xét tứ giác ODAE có

góc ODA+góc OEA=180 độ

=>ODAE là tứ giác nội tiếp

b: \(AE=\sqrt{\left(3R\right)^2-R^2}=2\sqrt{2}\cdot R\)

\(OI=\dfrac{OE^2}{OA}=\dfrac{R^2}{3R}=\dfrac{R}{3}\)

c: Xét ΔDIK vuông tại I và ΔDHE vuông tại H có

góc IDK chung

=>ΔDIK đồng dạng vơi ΔDHE

=>DI/DH=DK/DE

=>DH*DK=DI*DE=2*IE^2

Ta có: 

\(a=\sqrt[3]{7+\sqrt{50}}=\sqrt[3]{7+5\sqrt{2}}=\sqrt[3]{\left(1+\sqrt{2}\right)^3}=1+\sqrt{2}\)

\(b=\sqrt[3]{7-\sqrt{50}}=\sqrt[3]{7-5\sqrt{2}}=\sqrt[3]{\left(1-\sqrt{2}\right)^3}=1-\sqrt{2}\)

\(\Rightarrow M=a+b=2\) là số chẵn (đpcm)

Lại có:

\(a+b=2;a.b=\left(1+\sqrt{2}\right).\left(1-\sqrt{2}\right)=-1\)\(;a^2+b^2=\left(a+b\right)^2-2ab=6\)

\(N=a^7+b^7\)

\(=\left(a^7+a^4b^3\right)+\left(b^7+a^3b^4\right)-\left(a^4b^3+a^3b^4\right)\)

\(=a^4\left(a^3+b^3\right)+b^4\left(a^3+b^3\right)-a^3b^3\left(a+b\right)\)

\(=\left(a^3+b^3\right)\left(a^4+b^4\right)+2\)

\(=\left(a+b\right)\left(a^2+b^2-ab\right)\left[\left(a^2+b^2\right)^2-2a^2b^2\right]+2\)

\(=2\left(7.34+1\right)=478\) là số chẵn(đpcm)

ĐKXĐ : \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)

\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\div\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{x-x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}\div\frac{x-4-x+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}\times\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=\frac{x+2\sqrt{x}+1}{\sqrt{x}}\)

\(\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-\sqrt{x}-1}{x-2\sqrt{x}}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{x-5}{x-\sqrt{x}-2}\right)\)

\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}:\left(\frac{x-4-x+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)

\(=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)(p/s: sai thì mong bạn thông cảm nha)

\(\text{Đặt: }\sqrt{6+\sqrt{6+\sqrt{6+....}}}=a\Rightarrow a^2=6+a\Leftrightarrow a^2-a-6=\left(a-3\right)\left(a+2\right)=0\)

thấy ngay a không thể đạt giá trị âm nên 

a=3 thay vào P=0 (vô lí) -> đề sai.

a) Ta có:

\(A=\frac{\sqrt{x}-3}{x-\sqrt{x}+1}\)

\(A=\frac{\sqrt{4}-3}{4-\sqrt{4}+1}\)

\(A=\frac{2-3}{4-2+1}=-\frac{1}{3}\)

b) đk: \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

\(B=\left(\frac{3\sqrt{x}+6}{x-9}-\frac{2}{\sqrt{x}-3}\right):\frac{1}{\sqrt{x}+3}\)

\(B=\frac{3\sqrt{x}+6-2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\left(\sqrt{x}+3\right)\)

\(B=\frac{3\sqrt{x}+6-2\sqrt{x}-6}{\sqrt{x}-3}\)

\(B=\frac{\sqrt{x}}{\sqrt{x}-3}\)