\(x^2-3x+2\sqrt{x-3}=0\left(x\ge3\right)\\ \Leftrightarrow x\left(x-3\right)+2\sqrt{x-3}=0\)

Đặt \(x-3=t\)

\(\Leftrightarrow2t^2+xt=0\\ \Leftrightarrow t\left(2t+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=0\\2t=-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x-6=-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(N\right)\\x=2\left(L\right)\end{matrix}\right.\)

Ta có AB//CD 

 \(\Rightarrow\widehat{DAB}+\widehat{ADC}=180\\ \Rightarrow\widehat{ADC}+135=180\\ \Rightarrow\widehat{ADC}=45\)

Ta có \(\sin D=\sin45=\dfrac{AH}{AD}=\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\dfrac{AH}{15}=\dfrac{\sqrt{2}}{2}\left(cm\right)\\ \Rightarrow AH=\dfrac{15\sqrt{2}}{2}\left(cm\right)\\ \Rightarrow S_{ABCD}=AB\cdot AH=18\cdot\dfrac{15\sqrt{2}}{2}=135\left(cm^2\right)\)

Đặt đề hơi ảo vì có 2 góc H nên mình sẽ để CO cắt AB tại I

a)Xét tam giác ABC có \(\dfrac{BE}{AB}=\dfrac{CF}{AC}\Rightarrow EF//BC\Rightarrow EF\perp AH\)

Chứng minh được tam giác BEH = tam giác CFH (g.c.g)

\(\Rightarrow EH=HF\)

Nên E đx với F qua H

b) Ta có \(AH\cap BK\cap CI=O\)

Mà \(O\in AH\) và \(AH\) là đường cao

\(\Rightarrow\)BK và CI là đường cao 

Chứng minh được \(\Delta AKB=\Delta AIC\left(ch-gn\right)\)

\(\Rightarrow BK=CI;\widehat{ABK}=\widehat{ACI}\)

Mà BE=CF

\(\Rightarrow\Delta BEK=\Delta CFI\left(c.g.c\right)\)

\(\Rightarrow EK=FI\)

\(\dfrac{\sqrt[4]{86-14\sqrt{37}}}{\sqrt{9-2\sqrt{4+2\sqrt{3}}}-\sqrt{6+\sqrt{13+4\sqrt{3}}}}\)

\(=\dfrac{\sqrt{7-\sqrt{37}}}{\sqrt{9-2\left(\sqrt{3}+1\right)}-\sqrt{6+\left(2\sqrt{3}+1\right)}}\)

\(=\dfrac{\sqrt{7-\sqrt{37}}}{\sqrt{7-2\sqrt{3}}-\sqrt{7+2\sqrt{3}}}\)

\(=\dfrac{\sqrt{7-\sqrt{37}}\cdot\left(\sqrt{7-2\sqrt{3}}+\sqrt{7+2\sqrt{3}}\right)}{7-2\sqrt{3}-7-2\sqrt{3}}\)

\(=-\dfrac{\sqrt{7-\sqrt{37}}\cdot\left(\sqrt{7-2\sqrt{3}}+\sqrt{7+2\sqrt{3}}\right)}{4\sqrt{3}}\)

Tới đây chịu nha

12.

a)

\(P=\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\left(x>0;x\ne1\right)\\ P=\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)

Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

\(\Leftrightarrow A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)

\(\Leftrightarrow A^2=8+2\sqrt{16-10-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\sqrt{6-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\left(\sqrt{5}-1\right)\\ \Leftrightarrow A^2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\\ \Leftrightarrow A=\sqrt{5}+1\)

Vậy \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)

Bạn có thể lấy đó làm phần thân bài và tự triển khai ý của MB và KB vì nó dễ mak :))

2.

a)\(\sqrt{x+2}=4-x\left(x\ge-2\right)\)

\(\Leftrightarrow x+2=\left(4-x\right)^2\)

\(\Leftrightarrow x+2=16-8x+x^2\\ \Leftrightarrow x^2-9x+14=0\\ \Leftrightarrow\left(x-7\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\left(N\right)\\x=2\left(N\right)\end{matrix}\right.\)

b)\(\sqrt{x^2+1}=5-x^2\left(x\in R\right)\)

\(\Leftrightarrow x^2+1=25-10x^2+x^4\\ \Leftrightarrow x^4-11x^2+24=0\\ \Leftrightarrow\left(x^2-8\right)\left(x^2-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=8\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=-2\sqrt{2}\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

c)\(x^2+\sqrt{x^2-3x+5}=3x+7\left(x\in R\right)\)

\(\Leftrightarrow\left(x^2-3x+5\right)+\sqrt{x^2-3x+5}-12=0\)

Đặt \(\sqrt{x^2-3x+5}=t\left(t\ge\sqrt{\dfrac{11}{4}}\right)\)

\(\Leftrightarrow t+\sqrt{t}-12=0\\ \Leftrightarrow\left[{}\begin{matrix}t=-4\left(L\right)\\t=3\left(N\right)\end{matrix}\right.\\ \Leftrightarrow\sqrt{x^2+3x+5}=3\\ \Leftrightarrow x^2+3x+2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

d)\(\sqrt{x+2}-\sqrt{x-6}=2\left(x\ge6\right)\)

\(\Leftrightarrow x+2+x-6-2\sqrt{\left(x+2\right)\left(x-6\right)}=4\\ \Leftrightarrow2x-8-2\sqrt{\left(x+2\right)\left(x-6\right)}=0\\ \Leftrightarrow2\sqrt{\left(x+2\right)\left(x-6\right)}=2x-8\\ \Leftrightarrow\sqrt{\left(x+2\right)\left(x-6\right)}=x-4\\ \Leftrightarrow\left(x+2\right)\left(x-6\right)=x^2-8x+16\\ \Leftrightarrow x^2-4x-12=x^2-8x+16\\ \Leftrightarrow4x=28\Leftrightarrow x=7\left(N\right)\)

Tick nha

1.

a)\(\sqrt{300}+\sqrt{243}-\sqrt{192}=10\sqrt{3}+9\sqrt{3}-8\sqrt{3}=11\sqrt{3}\)

b)\(\sqrt{75}-\sqrt{48}+\sqrt{147}=5\sqrt{3}-4\sqrt{3}+7\sqrt{3}=8\sqrt{3}\)

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

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

\(=2\sqrt{2}-2+\dfrac{3\sqrt{2}+9}{7}+2-\sqrt{3}\)

\(=\dfrac{14\sqrt{2}-7\sqrt{3}+3\sqrt{2}+9}{7}=\dfrac{17\sqrt{2}-7\sqrt{3}+9}{7}\)

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