1.236067977

1.2360679777

\(\sqrt{6-2\sqrt{5}}=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)do \(\sqrt{5}-1>0\)

\(\frac{2+\sqrt{2}}{1+\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)

\(=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}+2-\sqrt{2}\)do \(2-\sqrt{2}>0\)

\(=\sqrt{2}+2-\sqrt{2}=2\)

a2+b2+c2=4−abc≤4a2+b2+c2=4−abc≤4

Smax=4Smax=4 khi 1 trong 3 số bằng 0

4=abc+a2+b2+c2≥abc+33√(abc)24=abc+a2+b2+c2≥abc+3(abc)23

Đặt 3√abc=x>0⇒x3+3x2−4≤0abc3=x>0⇒x3+3x2−4≤0

⇔(x−1)(x+2)2≤0⇒x≤1⇔(x−1)(x+2)2≤0⇒x≤1

⇒abc≤1⇒S=4−abc≥3⇒abc≤1⇒S=4−abc≥3

Dấu "=" xảy ra khi a=b=c=1

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)