Bài 1:
\(a,=5\sqrt{5}+15\sqrt{3}-4\sqrt{5}=15\sqrt{3}+\sqrt{5}\\ b,=\sqrt{2}+1-\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}+1-\sqrt{2}+1=2\\ c,=\dfrac{10-2\sqrt{2}+10+2\sqrt{2}}{\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)}=\dfrac{20}{21}\\ d,=\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}-\dfrac{5\left(\sqrt{2}+1\right)}{1}=\sqrt{2}-5\sqrt{2}-5=-4\sqrt{2}-5\)
Bài 2:
\(a,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=9\Leftrightarrow2x=8\Leftrightarrow x=4\left(tm\right)\\ b,ĐK:x\ge2\\ PT\Leftrightarrow x-2=2x+4\\ \Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow x\in\varnothing\)
Bài 3:
Áp dụng PTG: \(BC=\sqrt{AB^2+AC^2}=10\left(cm\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=3,6\left(cm\right)\\AH=\dfrac{AB\cdot AC}{BC}=4,8\left(cm\right)\end{matrix}\right.\)
