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\(ĐK:x\ne0;2\)
\(\Rightarrow\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)
\(\Leftrightarrow x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) 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
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
1. are visiting
2. don't go
3. are not going to work
4. Does Mary have
5. meet
are visting
don't go
won't work
Does Mary have
meet