Gọi BC,MN,AD cắt EF lần lượt tại P,P',G; AD cắt MN tại H

Ta có ngay \(\left(PDBC\right)=A\left(PDBC\right)=\left(PGFE\right)=-1\) và \(\left(P'GFE\right)=-1\)

Suy ra \(P\equiv P'\). Nói cách khác MN,EF,BC đồng quy tại P

Vì \(A\left(PDBC\right)=-1\), AD là phân giác \(\widehat{BAC}\) nên \(AD\perp AP\)

Lại do \(A\left(PHNM\right)=\left(PHNM\right)=\left(PGFE\right)=-1\) nên AH là phân giác \(\widehat{MAN}\).

Vô số nghiệm bạn ạ:))

\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{35}}\)

\(=\frac{2}{2\sqrt{1}}+\frac{2}{2\sqrt{2}}+...+\frac{2}{2\sqrt{35}}\)

\(>\frac{2}{\sqrt{1}+\sqrt{2}}+\frac{2}{\sqrt{2}+\sqrt{3}}+...+\frac{2}{\sqrt{35}+\sqrt{36}}\)

\(=2\left(\frac{2-1}{\sqrt{1}+\sqrt{2}}+\frac{3-2}{\sqrt{2}+\sqrt{3}}+...+\frac{36-35}{\sqrt{35}+\sqrt{36}}\right)\)

\(=2\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{36}-\sqrt{35}\right)\)

\(=2\left(\sqrt{36}-\sqrt{1}\right)\)

\(=2\left(6-1\right)=10\)

\(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}\)

\(=\sqrt{13+6\sqrt{4+\sqrt{\left(2\sqrt{2}-1\right)^2}}}\)

\(=\sqrt{13+6\sqrt{4+\sqrt{\left|2\sqrt{2}-1\right|}}}\)

\(=\sqrt{13+6\sqrt{4+2\sqrt{2}-1}}\)(do \(2\sqrt{2}>1\))

\(=\sqrt{13+6\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{13+6\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+6\left|\sqrt{2}+1\right|}\)

\(=\sqrt{13+6\left(\sqrt{2}+1\right)}\)(do \(\sqrt{2};1>0\))

\(=\sqrt{13+6\sqrt{2}+6}\)

\(=\sqrt{19+6\sqrt{2}}\)

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

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

\(=3\sqrt{2}+1\)(do \(3\sqrt{2};1>0\))

Đk: x \(\ge\)0

Với x = 0 => P = 0 (tm) => x =  0 thì P nhận giá trị nguyên

Ta có: P = \(\frac{2\sqrt{x}}{x-\sqrt{x}+1}=\frac{2}{\sqrt{x}-1+\frac{1}{\sqrt{x}}}\)

Để P nhận giá trị nguyên <=> \(2⋮\)\(\sqrt{x}-1+\frac{1}{\sqrt{x}}\)

<=> \(\sqrt{x}+\frac{1}{\sqrt{x}}-1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

Do \(\sqrt{x}+\frac{1}{\sqrt{x}}-1\ge2\cdot\sqrt{\sqrt{x}\cdot\frac{1}{\sqrt{x}}}-1=2-1=1\)

=> \(\sqrt{x}+\frac{1}{\sqrt{x}}-1\in\left\{1;2\right\}\)

Với \(\sqrt{x}+\frac{1}{\sqrt{x}}-1=1\) => \(\sqrt{x}+\frac{1}{\sqrt{x}}-2=0\) => \(\left(\sqrt{x}-1\right)^2=0\) <=> x = 1 (Tm)

\(\sqrt{x}+\frac{1}{\sqrt{x}}-1=2\) => \(x-3\sqrt{x}+1=0\)

Đặt \(\sqrt{x}=y\)=> pt trở thành y² - 3y + 1 = 0

\(\Delta=\left(-3\right)^2-4=5>0\) => pt có 2 nghiệm pb

\(y_1=\frac{3+\sqrt{5}}{2}\); \(y_2=\frac{3-\sqrt{5}}{2}\) 

\(y=\frac{3+\sqrt{5}}{2}\)=> \(\sqrt{x}=\frac{3+\sqrt{5}}{2}\)=> \(x=\left(\frac{3+\sqrt{5}}{2}\right)^2=\frac{14+6\sqrt{5}}{4}=\frac{7+3\sqrt{5}}{2}\)

 \(y=\frac{3-\sqrt{5}}{2}\) => \(\sqrt{x}=\frac{3-\sqrt{5}}{2}\)=> \(x=\frac{7-3\sqrt{5}}{2}\)

Ta có: \(x=\frac{3-\sqrt{5}}{2}=\frac{6-2\sqrt{5}}{4}=\frac{\left(\sqrt{5}-1\right)^2}{4}\)=> \(\sqrt{x}=\sqrt{\frac{\left(\sqrt{5}-1\right)^2}{4}}=\frac{\sqrt{5}-1}{2}\)

Do đó: P = \(\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{\frac{\sqrt{5}-1}{2}+1}{\frac{\sqrt{5}-1}{2}-2}=\frac{\frac{\sqrt{5}+1}{2}}{\frac{\sqrt{5}-5}{2}}=\frac{\sqrt{5}+1}{\sqrt{5}-5}\)

\(P=\frac{\left(\sqrt{5}+1\right)\left(5+\sqrt{5}\right)}{5-25}=\frac{6\sqrt{5}+10}{-20}=-\frac{3\sqrt{5}+5}{10}\)

k cho mik

kb nữa nha

thanks

hok tốt