Đk: \(x\ne\pm1\)

\(\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\right).\dfrac{4x^2-4}{5}\)

\(=\left[\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right].\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{\left(x+1\right)^2+6-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}.\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(=\dfrac{10}{2\left(x-1\right)\left(x+1\right)}.\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)\(=4\)

\(a.ĐK:x\ge0;x\ne1\)

\(M=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}+\dfrac{1}{x+1}\right].\dfrac{x+1}{\sqrt{x}-1}\)

\(=\left[\dfrac{\sqrt{x}}{x+1}+\dfrac{1}{x+1}\right].\dfrac{x+1}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}+1}{x+1}.\dfrac{x+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b)\(M\sqrt{x}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)+2}{\sqrt{x}-1}\)

\(=\sqrt{x}+2+\dfrac{2}{\sqrt{x}-1}=\left(\sqrt{x}-1+\dfrac{2}{\sqrt{x}-1}\right)+3\ge2\sqrt{\left(\sqrt{x}-1\right).\dfrac{2}{\sqrt{x}-1}}+3=2\sqrt{2}+3\)

\(\Rightarrow M\sqrt{x}\ge2\sqrt{2}+3\)

Dấu "=" xảy ra khi \(\sqrt{x}-1=\dfrac{2}{\sqrt{x}-1}\Leftrightarrow\sqrt{x}=\sqrt{2}+1\) (tm)

Vậy GTNN của \(M\sqrt{x}=2\sqrt{2}+3\)

1.C

2.A

3.B

4.C

5.C

6.C

7.B

8.A

\(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}\)\(=\sqrt{\dfrac{6-2\sqrt{5}}{6+2\sqrt{5}}}-\sqrt{\dfrac{6+2\sqrt{5}}{6-2\sqrt{5}}}\)

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

\(=\dfrac{\left|\sqrt{5}-1\right|}{\sqrt{5}+1}-\dfrac{\sqrt{5}+1}{\left|\sqrt{5}-1\right|}\)

\(=\dfrac{\sqrt{5}-1}{\sqrt{5}+1}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2-\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=-\dfrac{4\sqrt{5}}{4}\)\(=-\sqrt{5}\)

Vậy...

Có \(x+y=7+4\sqrt{3}+7-4\sqrt{3}=14\)

\(xy=\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)=1\)

\(x^2+y^2=\left(x+y\right)^2-2xy=14^2-2=194\)

\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=14^3-3.1.14=2702\)

\(x^7+y^7=\left(x^3+y^3\right)\left(x^4+y^4\right)-x^3y^3\left(x+y\right)\)\(=2702\left[\left(x^2+y^2\right)^2-2x^2y^2\right]-14\)

\(=2702\left(194^2-2\right)-14=101687054\)

Vậy...

a)Áp dụng định lí py-ta-go có:

 \(DE=\sqrt{OD^2+OE^2}=\sqrt{R^2+R^2}=\sqrt{2}R\)

Dễ chứng minh được: \(\Delta EBC\sim\Delta DAC\left(g.g\right)\)

\(\Rightarrow\dfrac{BC}{AC}=\dfrac{CE}{DC}\)\(\Rightarrow CD=\dfrac{AC.BC}{EC}=\dfrac{\left(OA+OC\right).\left(OC-OB\right)}{DC-DE}\)

\(\Leftrightarrow CD=\dfrac{8R^2}{DC-\sqrt{2}R}\)

\(\Leftrightarrow DC^2-\sqrt{2}R.DC-8R^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}CD=\dfrac{R\left(\sqrt{34}+\sqrt{2}\right)}{2}\\CD=\dfrac{R\left(-\sqrt{34}+\sqrt{2}\right)}{2}\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow CD=\dfrac{R\left(\sqrt{34}+\sqrt{2}\right)}{2}\)

Có \(EC=DC-DE=\dfrac{R\left(\sqrt{34}+\sqrt{2}\right)}{2}-\sqrt{2}R=\dfrac{R\left(\sqrt{34}-\sqrt{2}\right)}{2}\)

Vậy...

22:A

23:B

24:A