\(ĐK:x\ne0;2\)

\(\Rightarrow\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

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

\(\Leftrightarrow x^2+2x-x+2-2=0\)

\(\Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\)       ĐKXĐ: x≠0, x≠2

<=>\(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)

=>x(x+2) - x + 2 =2

<=>x\(^2\) + 2x - x = 2-2

<=> x\(^2\) + x =0

<=> x(x + 1) = 0

<=>\(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) 

mà ĐKXĐ: x≠0, x≠2

Vậy x = -1

\(A=\sqrt{3-\sqrt{5}}-\sqrt{4-\sqrt{15}}+\sqrt{6-3\sqrt{3}}\)

\(=\dfrac{1}{\sqrt{2}}\cdot\left(\sqrt{6-2\sqrt{5}}-\sqrt{8-2\sqrt{15}}+\sqrt{12-6\sqrt{3}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{5}-1-\sqrt{5}+\sqrt{3}+3-\sqrt{3}\right)\)

=2/căn 2=căn 2

\(B=\sqrt{4-\sqrt{7}}-\sqrt{14-5\sqrt{3}}-\sqrt{5+\sqrt{21}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{7}}-\sqrt{28-10\sqrt{3}}-\sqrt{10+2\sqrt{21}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}-1-5+\sqrt{3}-\sqrt{7}-\sqrt{3}\right)\)

=-6/căn 2=-3căn2

\(C=\sqrt{11-6\sqrt{2}}-\sqrt{6-4\sqrt{2}}+\sqrt{7-2\sqrt{6}}\)

=3-căn 2-2+căn 2+căn 6-1

=căn 6

\(D=\sqrt{6-\sqrt{11}}-\sqrt{10+3\sqrt{11}}+2\sqrt{2}-1\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{12-2\sqrt{11}}-\sqrt{20+6\sqrt{11}}\right)+2\sqrt{2}-1\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{11}-1-\sqrt{11}-3\right)+2\sqrt{2}-1\)

=-1

\(F=\sqrt{6+3\sqrt{3}}-\sqrt{2+\sqrt{3}}+\sqrt{6-4\sqrt{2}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{12+6\sqrt{3}}-\sqrt{4+2\sqrt{3}}\right)+2-\sqrt{2}\)

=1/căn 2(3+căn 3-căn 3-1)+2-căn 2

=căn 2+2-căn 2

=2

15:

1:

góc A=2*góc B

mà góc A+góc B=180-60=120 độ

nên góc A=80 độ; góc B=40 độ

2: 

a: Xet ΔAMB và ΔAMC có

AM chung

MB=MC

AB=AC

=>ΔAMB=ΔAMC

b: Xét ΔMBD và ΔMCE có

MB=MC

góc MBD=góc MCE

BD=CE

=>ΔMBD=ΔMCE

=>MD=ME

c: ΔADE cân tại A

mà AN là phân giác

nên AN vuông góc DE

  chọn D

đk : x khác 1 ; -3 

\(\Rightarrow2x+6+4x-4=3x+11\Leftrightarrow3x=9\Leftrightarrow x=3\left(tm\right)\)

\(a))\Delta ABC:\)

\(\dfrac{AM}{AB}=\dfrac{AN}{AC}\left(do\dfrac{4}{12}=\dfrac{5}{15}\right).\\ \Rightarrow MN//BC\left(Talet\right).\)

\(b))\Delta ABC:MN//BC\left(cmt\right).\\ \Rightarrow\dfrac{MN}{BC}=\dfrac{AM}{AB}\left(Talet\right).\\ \Rightarrow MN=\dfrac{AM.BC}{AB}=\dfrac{4.21}{12}=7\left(cm\right).\)

2:

A=(x1-x2)^2-x1^2+x1(x1+x2)

=(x1-x2)^2+x1x2

=(x1+x2)^2-3x1x2

=(1/2)^2-3*(-1/4)=1/4+3/4=1