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Gọi O là tâm đường tròn \(\Rightarrow\) O là trung điểm BC
\(\stackrel\frown{BE}=\stackrel\frown{ED}=\stackrel\frown{DC}\Rightarrow\widehat{BOE}=\widehat{EOD}=\widehat{DOC}=\dfrac{180^0}{3}=60^0\)
Mà \(OD=OE=R\Rightarrow\Delta ODE\) đều
\(\Rightarrow ED=R\)
\(BN=NM=MC=\dfrac{2R}{3}\Rightarrow\dfrac{NM}{ED}=\dfrac{2}{3}\)
\(\stackrel\frown{BE}=\stackrel\frown{DC}\Rightarrow ED||BC\)
Áp dụng định lý talet:
\(\dfrac{AN}{AE}=\dfrac{MN}{ED}=\dfrac{2}{3}\Rightarrow\dfrac{EN}{AN}=\dfrac{1}{2}\)
\(\dfrac{ON}{BN}=\dfrac{OB-BN}{BN}=\dfrac{R-\dfrac{2R}{3}}{\dfrac{2R}{3}}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{EN}{AN}=\dfrac{ON}{BN}=\dfrac{1}{2}\) và \(\widehat{ENO}=\widehat{ANB}\) (đối đỉnh)
\(\Rightarrow\Delta ENO\sim ANB\left(c.g.c\right)\)
\(\Rightarrow\widehat{NBA}=\widehat{NOE}=60^0\)
Hoàn toàn tương tự, ta có \(\Delta MDO\sim\Delta MAC\Rightarrow\widehat{MCA}=\widehat{MOD}=60^0\)
\(\Rightarrow\Delta ABC\) đều
a, \(\hept{\begin{cases}x^2+y^2+3xy=5\\\left(x+y\right)\left(x+y+1\right)+xy=7\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2+xy=5\\\left(x+y\right)\left(x+y+1\right)+xy=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2-\left(x+y\right)\left(x+y+1\right)=-2\\\left(x+y\right)^2+xy=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+y\right)\left(x+y-x-y-1\right)=-2\\\left(x+y\right)^2+xy=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=2\\4+xy=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2-y\\4+\left(2-y\right)y=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2-y\\2y-y^2-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2-y\\-\left(y^2-2y+1\right)=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2-y\\\left(y-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}}\)
Vậy hpt có nghiệm (x;y) = (1;1)
cho mik lời giải nữa nhá thanks
Mọi người giúp em bài này với ạ.EM cần gấp ạ
Mọi người giúp em bài này với ạ.EM cần gấp ạ
Chứng minh 2 ý đó giúp e vs ạ e đang cần gấp
ta có :
\(\frac{1}{cos^2x}=\frac{sin^2x+cos^2x}{cos^2x}=1+\left(\frac{sinx}{cosx}\right)^2=1+tan^2x\)
\(\frac{1}{sin^2x}=\frac{sin^2x+cos^2x}{sin^2x}=1+\left(\frac{cosx}{sinx}\right)^2=1+cot^2x\)
29: Ta có: \(\dfrac{1}{\sqrt{7}+\sqrt{5}}+\dfrac{2}{1-\sqrt{7}}\)
\(=\dfrac{\sqrt{7}-\sqrt{5}}{2}-\dfrac{2\sqrt{7}-2}{6}\)
\(=\dfrac{3\sqrt{7}-3\sqrt{5}-2\sqrt{7}+2}{6}\)
\(=\dfrac{-3\sqrt{5}-2}{6}\)
30: Ta có: \(\dfrac{4}{1-\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
\(=\dfrac{-4\sqrt{3}-4}{2}+\dfrac{4-2\sqrt{3}}{2}\)
\(=\dfrac{-4\sqrt{3}-4+4-2\sqrt{3}}{2}=-3\sqrt{3}\)
31: Ta có: \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
\(=-\sqrt{3}-\sqrt{2}-\dfrac{3}{3\sqrt{2}+2\sqrt{3}}\)
\(=-\sqrt{3}-\sqrt{2}-\dfrac{9\sqrt{2}-6\sqrt{3}}{6}\)
\(=\dfrac{-6\sqrt{3}-6\sqrt{2}-9\sqrt{2}+6\sqrt{3}}{6}=\dfrac{-15\sqrt{2}}{6}\)
\(=\dfrac{-5\sqrt{2}}{2}\)
29.
\(=\frac{\sqrt{7}-\sqrt{5}}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})}+\frac{2(1+\sqrt{7})}{(1-\sqrt{7})(1+\sqrt{7})}\)
\(=\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{2(1+\sqrt{7})}{1-7}=\frac{\sqrt{7}-\sqrt{5}}{2}-\frac{1+\sqrt{7}}{3}=\frac{\sqrt{7}-3\sqrt{5}-2}{6}\)
30.
\(=\frac{4(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}+\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{4(1+\sqrt{3})}{1-3}+\frac{4-2\sqrt{3}}{3-1}\)
\(=\frac{4(1+\sqrt{3})}{-2}+\frac{4-2\sqrt{3}}{2}=-2(1+\sqrt{3})+2-\sqrt{3}=-3\sqrt{3}\)
31.
\(=\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}-\frac{3(\sqrt{18}-2\sqrt{3})}{(\sqrt{18}+2\sqrt{3})(18-2\sqrt{3})}=\frac{\sqrt{2}+\sqrt{3}}{2-3}-\frac{3(18-2\sqrt{3})}{6}\)
\(=-(\sqrt{2}+\sqrt{3})-(9-\sqrt{3})=-\sqrt{2}-9\)
32.
\(=\frac{(\sqrt{2}-1)^2}{(\sqrt{2}-1)(\sqrt{2}+1)}-\frac{(3-\sqrt{2})^2}{(3-\sqrt{2})(3+\sqrt{2})}=\frac{3-2\sqrt{2}}{2-1}-\frac{11-6\sqrt{2}}{3^2-2}\)
\(=3-2\sqrt{2}-\frac{11-6\sqrt{2}}{7}=\frac{10-8\sqrt{2}}{7}\)
33.
\(=\frac{(\sqrt{5}+\sqrt{6})^2+(\sqrt{6}-\sqrt{5})(\sqrt{5}-\sqrt{6})}{(\sqrt{5}-\sqrt{6})(\sqrt{6}+\sqrt{5})}=\frac{4\sqrt{30}}{5-6}=-4\sqrt{30}\)
34.
\(=\frac{3(1+\sqrt{2})}{(1-\sqrt{2})(1+\sqrt{2})}+\frac{(\sqrt{2}-1)^2}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{3(1+\sqrt{2})}{1-2}+\frac{3-2\sqrt{2}}{2-1}\)
\(=-3(1+\sqrt{2})+3-2\sqrt{2}=-5\sqrt{2}\)
35.
\(=\frac{(\sqrt{5}-1)(1-\sqrt{5})+6(\sqrt{5}+1)}{(\sqrt{5}+1)(1-\sqrt{5})}=\frac{2\sqrt{5}-6+6\sqrt{5}+6}{1-5}=\frac{8\sqrt{5}}{-4}=-2\sqrt{5}\)
36.
\(=\frac{(\sqrt{2}+\sqrt{3})(\sqrt{6}+2)+(\sqrt{3}-\sqrt{2})(2-\sqrt{6})}{(2-\sqrt{6})(\sqrt{6}+2)}\)
\(=\frac{8\sqrt{3}}{2^2-6}=\frac{8\sqrt{3}}{-2}=-4\sqrt{3}\)