a) Vì ADHE nội tiếp \(\Rightarrow\angle AED=\angle AHD=90-\angle BHD=\angle DBH\)

\(\Rightarrow BDEC\) nội tiếp

b) Xét \(\Delta SCE\) và \(\Delta SDB:\) Ta có: \(\left\{{}\begin{matrix}\angle SCE=\angle SDB\\\angle DSBchung\end{matrix}\right.\)

\(\Rightarrow\Delta SCE\sim\Delta SDB\left(g-g\right)\Rightarrow\dfrac{SE}{SC}=\dfrac{SB}{SD}\Rightarrow SE.SD=SB.SC\left(1\right)\)

Ta có: \(\left\{{}\begin{matrix}SH\bot HO\\H\in\left(O\right)\end{matrix}\right.\Rightarrow\) SH là tiếp tuyến của (O) 

Xét \(\Delta SHE\) và \(\Delta SDH:\) Ta có: \(\left\{{}\begin{matrix}\angle SHE=\angle SDH\\\angle DSHchung\end{matrix}\right.\)

\(\Rightarrow\Delta SHE\sim\Delta SDH\Rightarrow\dfrac{SH}{SE}=\dfrac{SD}{SH}\Rightarrow SE.SD=SH^2\left(2\right)\)

Từ (1) và (2) \(\Rightarrow SH^2=SB.SC\)

1.1) \(\left\{{}\begin{matrix}2x-3y=-7\\4x+y=-7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x-6y=-14\left(1\right)\\4x+y=-7\left(2\right)\end{matrix}\right.\)

Lấy \(\left(2\right)-\left(1\right)\),ta được: \(7y=7\Rightarrow y=1\Rightarrow2x=-7+3=-4\Rightarrow x=-2\)

2) \(P=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\right).\dfrac{\sqrt{x}}{2x-\sqrt{x}-3}\)

\(=\left(\dfrac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}\right).\dfrac{\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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

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

\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}.\dfrac{\sqrt{x}}{\left(2x-3\right)\left(x+1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}.\dfrac{\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}-3}\)

Gọi số quả cam của rổ 1 và 2 lần lượt là a,b(quả) \(\left(a,b>6\right)\)

Theo đề: \(\left\{{}\begin{matrix}a=b\\b+6=\left(a-6\right)^2\end{matrix}\right.\)

\(\Rightarrow a+6=\left(a-6\right)^2\Leftrightarrow a^2-13a+30=0\Leftrightarrow\left(a-3\right)\left(a-10\right)=0\)

mà \(a>6\Rightarrow a=10\Rightarrow a=b=10\) \(\)

ĐKXĐ: \(x>0;x\ne1,4\)

\(\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)

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

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

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

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

1) Ta có: \(\angle BAC=\angle ACD=\angle ABD=90\Rightarrow ABDC\) là hình chữ nhật

Có I là giao điểm của 2 đường chéo \(\Rightarrow\) I là trung điểm BC

mà \(\Delta ABC\) vuông tại A \(\Rightarrow AI=\dfrac{1}{2}BC\)

2) Ta có: \(\angle HAI=90-\angle HIA=90-2\angle ICA=90-\angle IAC-\angle ICA\)

\(=\angle BAI-\angle ACB=\angle ABC-\angle ACB\)

3) Ta có: \(\angle AMD+\angle ACD=90+90=180\Rightarrow AMDC\) nội tiếp

\(\Rightarrow\angle AMC=\angle ADC=\angle ABC=\angle HAC\Rightarrow\Delta CAM\) cân tại C

có \(CB\bot AM\Rightarrow CB\) là trung trực AM \(\Rightarrow BA=MB\Rightarrow\Delta MAB\) cân tại B

\(\Rightarrow\angle BMH=\angle BAH=90-\angle ABC=\angle ACB=\angle IAC=\angle CMD\)

\(\Rightarrow\angle BMC=\angle BMH+\angle AMC=\angle CMD+\angle AMC=\angle AMD=90\)

c) Vì MEIB nội tiếp \(\Rightarrow\angle MIE=\angle MBE=45=\angle IME\Rightarrow\Delta IEM\) vuông cân tại E

Ta có: \(\angle CEM+\angle BEM=\angle IEB+\angle BEM=90\) \(\Rightarrow\angle CEM=\angle IEB\)

Xét \(\Delta CEM\) và \(\Delta BEI:\) Ta có: \(\left\{{}\begin{matrix}CE=BE\\\angle CEM=\angle BEI\\EM=EI\end{matrix}\right.\)

\(\Rightarrow\Delta CEM=\Delta BEI\left(c-g-c\right)\Rightarrow CM=BI\Rightarrow AI=BM\)

Ta có: \(\dfrac{AM}{MN}=\dfrac{BM}{MC}=\dfrac{AI}{IB}\Rightarrow\) \(MI\parallel BK\Rightarrow \angle IMB=\angle MBK\)

mà \(\angle IMB=\angle IEB=\angle CEM\Rightarrow\) CEBK nội tiếp

mà \(\angle CEB=90\Rightarrow CKB=90\Rightarrow CK\bot BN\)

b) Ta có: \(\angle MEI+\angle MBI=90+90=180\Rightarrow\) MEIB nội tiếp

\(\Rightarrow\angle IME=\angle IBE=\angle ABD=45\)

Gọi vận tốc xe ô tô tải là a(km/h) \(\left(a>0\right)\), t là thời gian xe ô tô con bắt đầu đi tới khi đuổi kịp xe ô tô tải \(\left(t>0\right)\)

Theo đề,ta có: \(\dfrac{1}{2}.a+ta=t\left(a+20\right)=120\)

\(\Leftrightarrow\dfrac{1}{2}a=20t\Rightarrow a=40t\Rightarrow t\left(40t+20\right)=120\)

\(\Leftrightarrow40t^2+20t=120\Rightarrow2t^2+t-6=0\Leftrightarrow\left(t+2\right)\left(2t-3\right)=0\)

mà \(t>0\Rightarrow t=\dfrac{3}{2}\Rightarrow a=60\) (km/h)

\(\Rightarrow\) vận tốc xe ô tô con là \(60+20=80\) (km/h)

2) \(B=\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\sqrt{\dfrac{6-2\sqrt{5}}{2}}}{\sqrt{2}}-\dfrac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

\(=\dfrac{\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2}}}{\sqrt{2}}-\dfrac{6-2\sqrt{5}}{4}=\dfrac{\dfrac{\sqrt{5}-1}{\sqrt{2}}}{\sqrt{2}}-\dfrac{6-2\sqrt{5}}{4}\)

\(=\dfrac{\sqrt{5}-1}{2}-\dfrac{6-2\sqrt{5}}{4}=\dfrac{2\sqrt{5}-2-6+2\sqrt{5}}{4}=\sqrt{5}-2\)

3) \(x-3\sqrt{x}+2=0\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

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