Bài 5:
1: ΔAHD vuông tại H
=>\(AH^2+HD^2=AD^2\)
=>\(HD^2=4,5^2-4^2=0,5\cdot8,5=4,25\)
=>\(HD=\sqrt{4,25}=\sqrt{\frac{17}{4}}=\frac{\sqrt{17}}{2}\) (cm)
Xét ΔABD vuông tại A có AH là đường cao
nên \(AD^2=DH\cdot DB\)
=>\(DB=4,5^2:\frac{\sqrt{17}}{2}=20,25\cdot\frac{2}{\sqrt{17}}=\frac{40.5}{\sqrt{17}}=\frac{81}{2\sqrt{17}}\) (cm)
Diện tích tam giác ABD là:
\(S_{ABD}=\frac12\cdot AH\cdot BD=\frac12\cdot4\cdot\frac{81}{2\sqrt{17}}=\frac{81}{\sqrt{17}}\left(\operatorname{cm}^2\right)\)
ABCD là hình chữ nhật
=>\(S_{ABCD}=2\cdot S_{ABD}=2\cdot\frac{81}{\sqrt{17}}=\frac{162}{\sqrt{17}}\left(\operatorname{cm}^2\right)\)
Bài 4:
1: \(\frac{2\sqrt{x}+3\sqrt{y}}{\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6}\)
\(=\frac{2\sqrt{x}+3\sqrt{y}}{\sqrt{x}\left(\sqrt{y}+2\right)-3\left(\sqrt{y}+2\right)}=\frac{2\sqrt{x}+3\sqrt{y}}{\left(\sqrt{y}+2\right)\left(\sqrt{x}-3\right)}\)
\(\frac{6-\sqrt{xy}}{\sqrt{xy}+2\sqrt{x}+3\sqrt{y}+6}\)
\(=\frac{6-\sqrt{xy}}{\sqrt{x}\left(\sqrt{y}+2\right)+3\left(\sqrt{y}+2\right)}=\frac{6-\sqrt{xy}}{\left(\sqrt{y}+2\right)\left(\sqrt{x}+3\right)}\)
Ta có: \(M=\frac{2\sqrt{x}+3\sqrt{y}}{\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6}-\frac{6-\sqrt{xy}}{\sqrt{xy}+2\sqrt{x}+3\sqrt{y}+6}\)
\(=\frac{2\sqrt{x}+3\sqrt{y}}{\left(\sqrt{x}-3\right)\left(\sqrt{y}+2\right)}+\frac{\sqrt{xy}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{y}+2\right)}\)
\(=\frac{\left(2\sqrt{x}+3\sqrt{y}\right)\left(\sqrt{x}+3\right)+\left(\sqrt{xy}-6\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\cdot\left(\sqrt{y}+2\right)}\)
\(=\frac{2x+6\sqrt{x}+3\sqrt{xy}+9\sqrt{y}+x\sqrt{y}-3\sqrt{xy}-6\sqrt{x}+18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\cdot\left(\sqrt{y}+2\right)}=\frac{x\sqrt{y}+2x+9\sqrt{y}+18}{\left(x-9\right)\left(\sqrt{y}+2\right)}\)
\(=\frac{\left(x+9\right)\left(\sqrt{y}+2\right)}{\left(x-9\right)\left(\sqrt{y}+2\right)}=\frac{x+9}{x-9}\)
Khi x=2020 thì \(M=\frac{2020+9}{2020-9}=\frac{2029}{2011}\)
