AMBC nội tiếp \(\Rightarrow\angle AMB=\angle ACB\)

Vì tam giác ABC cân tại A có góc A = 40 

\(\Rightarrow\angle ACB=\dfrac{180-\angle BAC}{2}=\dfrac{180-40}{2}=70\)

\(\Rightarrow\angle AMB=70\)

a) \(\sqrt{20}-\sqrt{45}-\sqrt{180}+3\sqrt{405}=\sqrt{4.5}-\sqrt{9.5}-\sqrt{36.5}+3.\sqrt{81.5}\)

\(=2\sqrt{5}-3\sqrt{5}-6\sqrt{5}+3.9\sqrt{5}=2\sqrt{5}-3\sqrt{5}-6\sqrt{5}+27\sqrt{5}\)

\(=20\sqrt{5}\)

b) \(\left(\sqrt{6}-\sqrt{5}\right)^2+\sqrt{120}=6+5-2\sqrt{30}+\sqrt{4.30}\)

\(=11-2\sqrt{30}+2\sqrt{30}=11\)

c) \(\dfrac{1}{2}\sqrt{6}-\sqrt{\dfrac{3}{2}}+3\sqrt{\dfrac{2}{3}}=\dfrac{1}{2}\sqrt{6}-\sqrt{\dfrac{1}{4}.6}+\sqrt{9.\dfrac{2}{3}}\)

\(=\dfrac{1}{2}\sqrt{6}-\dfrac{1}{2}\sqrt{6}+\sqrt{6}=\sqrt{6}\)

a) \(\sqrt{10+4\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{6}\right)^2+2.\sqrt{6}.2+2^2}-\sqrt{\left(\sqrt{6}\right)^2-2.\sqrt{6}.2+2^2}\)

\(=\sqrt{\left(\sqrt{6}+2\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}+2\right|-\left|\sqrt{6}-2\right|\)

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

b) \(\sqrt{39-12\sqrt{3}}+\sqrt{39+12\sqrt{3}}\)

\(=\sqrt{6^2-2.6.\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{6^2+2.6.\sqrt{3}+\left(\sqrt{3}\right)^2}\)

\(=\sqrt{\left(6-\sqrt{3}\right)^2}+\sqrt{\left(6+\sqrt{3}\right)^2}=\left|6-\sqrt{3}\right|+\left|6+\sqrt{3}\right|\)

\(=6-\sqrt{3}+6+\sqrt{3}=12\)

c) \(\sqrt{31-12\sqrt{3}}-\sqrt{31+12\sqrt{3}}\)

\(=\sqrt{\left(3\sqrt{3}\right)^2-2.3\sqrt{3}.2+2^2}-\sqrt{\left(3\sqrt{3}\right)^2+2.3\sqrt{3}.2+2^2}\)

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

\(=3\sqrt{3}-2-3\sqrt{3}-2=-4\)

d) \(\sqrt{21+12\sqrt{3}}+\sqrt{21-12\sqrt{3}}\)

\(=\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}.3+3^2}+\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}.3+3^2}\)

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

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

\(\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}\)

\(=\sqrt{a-2+4\sqrt{a-2}+4}+\sqrt{a-2-4\sqrt{a-2}+4}\)

\(=\sqrt{\left(\sqrt{a-2}+2\right)^2}+\sqrt{\left(\sqrt{a-2}-2\right)^2}\)

\(=\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|\)

Vì \(a\le6\Rightarrow\sqrt{a-2}\le2\Rightarrow\sqrt{a-2}-2\le0\Rightarrow\left|\sqrt{a-2}-2\right|=2-\sqrt{a-2}\)

Vì \(a\ge2\Rightarrow\sqrt{a-2}+2\ge2>0\)

\(\Rightarrow\text{ }\left|\sqrt{a-2}+2\right|+\left|\sqrt{a-2}-2\right|=\sqrt{a-2}+2+2-\sqrt{a-2}=4\)

Kẻ \(AE,BF\bot CD\)

Vì \(AE\parallel BF(\bot CD),AB\parallel EF\) (ABCD là hình thang cân)

\(\Rightarrow ABFE\) là hình bình hành có \(\angle AEF=90\Rightarrow ABFE\) là hình chữ nhật

\(\Rightarrow AB=FE\)

Dễ dàng chứng minh được \(DE=CF\left(\Delta ADE=\Delta BFC\right)\)

\(\Rightarrow DE=\dfrac{CD-AB}{2}=\dfrac{7-3}{2}=2\)

\(\Rightarrow AE=\sqrt{AD^2-DE^2}=\sqrt{5^2-2^2}=\sqrt{21}\)

\(\Rightarrow S_{ABCD}=\dfrac{1}{2}\left(AB+CD\right).AE=\dfrac{1}{2}\left(7+3\right).\sqrt{21}=5\sqrt{21}\)

a) \(\dfrac{3-\sqrt{x}}{x-9}=\dfrac{3-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\dfrac{1}{\sqrt{x}+3}\)

b) \(\dfrac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\sqrt{x}-2\)

c) \(6-2x-\sqrt{9-6x+x^2}=6-2x-\sqrt{\left(3-x\right)^2}=6-2x-\left|3-x\right|\)

mà \(x< 3\Rightarrow3-x>0\Rightarrow6-2x-\left|3-x\right|=6-2x-3+x=3-x\)

\(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\Rightarrow\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Rightarrow2a^2+2b^2-a^2-2ab-b^2\ge0\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow\left(a-b\right)^2\ge0\) (luôn đúng) \(\Rightarrow\) đpcm

e) \(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\left(x\ge-2\right)\)

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

\(\Leftrightarrow3\left(1+\sqrt{\left(x+2\right)\left(x+5\right)}\right)=3\left(\sqrt{x+5}+\sqrt{x+2}\right)\)

\(\Rightarrow1+\sqrt{\left(x+2\right)\left(x+5\right)}=\sqrt{x+5}+\sqrt{x+2}\)

\(\Rightarrow\sqrt{x+5}+\sqrt{x+2}-\sqrt{\left(x+2\right)\left(x+5\right)}-1=0\)

\(\Leftrightarrow\left(1-\sqrt{x+5}\right)\left(\sqrt{x+2}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}1=\sqrt{x+5}\\\sqrt{x+2}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

mà \(x\ge-2\Rightarrow x=-1\)

a) Ta có: \(OA=OB=OC=R\Rightarrow\Delta ABC\) vuông tại A

\(\Rightarrow\angle BAC=90\)

b) Vì \(OA=OB\Rightarrow\Delta OAB\) cân tại O \(\Rightarrow\angle OAB=\angle OBA\)

mà \(\angle BAx=\angle BCA\Rightarrow\angle BAx+\angle BAO=\angle BCA+\angle ABO\)

\(\Rightarrow\angle OAx=90\Rightarrow Ax\bot AO\Rightarrow Ax\) là tiếp tuyến của (O)

a) Xét \(\Delta AEB\) và \(\Delta AFC:\) Ta có: \(\left\{{}\begin{matrix}\angle AEB=\angle AFC=90\\\angle BACchung\end{matrix}\right.\)

\(\Rightarrow\Delta AEB\sim\Delta AFC\left(g-g\right)\)

b) \(\Delta AEB\sim\Delta AFC\Rightarrow\dfrac{AE}{AF}=\dfrac{AB}{AC}\Rightarrow\dfrac{AE}{AB}=\dfrac{AF}{AC}\)

Xét \(\Delta AEF\) và \(\Delta ABC:\) Ta có: \(\left\{{}\begin{matrix}\dfrac{AE}{AB}=\dfrac{AF}{AC}\\\angle BACchung\end{matrix}\right.\)

\(\Rightarrow\Delta AEF\sim\Delta ABC\left(c-g-c\right)\)

c) Xét \(\Delta BFC\) và \(\Delta BDA:\) Ta có: \(\left\{{}\begin{matrix}\angle BFC=\angle BDA=90\\\angle ABCchung\end{matrix}\right.\)

\(\Rightarrow\Delta BFC\sim\Delta BDA\left(g-g\right)\Rightarrow\dfrac{BF}{BD}=\dfrac{BC}{BA}\Rightarrow BF.BA=BC.BD\)

Xét \(\Delta CEB\) và \(\Delta CDA:\) Ta có: \(\left\{{}\begin{matrix}\angle BEC=\angle CDA=90\\\angle ACBchung\end{matrix}\right.\)

\(\Rightarrow\Delta CEB\sim\Delta CDA\left(g-g\right)\Rightarrow\dfrac{CE}{CD}=\dfrac{CB}{CA}\Rightarrow CE.CA=CD.BC\)

\(\Rightarrow BF.BA+CE.CA=BC.BD+BC.CD=BC\left(BD+CD\right)=BC^2\)