nãy mình giải giúp hết rồi mà bạn 

bh=3;ch=12
=> bc=bh+ch=3+12=15(cm)
sử dụng hệ thức trong tam giác vuông
ah^2=bh.hc=36
=>ah=6(cm)
ab.ac=ah.bc=6.15=90
ab=90/ac
áp dụng định lí pitago
ab^2+ac^2=bc^2
<=> (90/ac)^2+ac^2=225
<=> 8100/ac^2+ac^2=225

 

\(3\sqrt{44}-2\sqrt{99}=3\sqrt{4.11}-2\sqrt{9.11}=6\sqrt{11}-6\sqrt{11}=0\)
\(\sqrt{72}-\sqrt{8}=\sqrt{9.8}-\sqrt{8}=3\sqrt{8}-\sqrt{8}=2\sqrt{8=4\sqrt{2}}\)
\(2\sqrt{8}-5\sqrt{18}+4\sqrt{32}=4\sqrt{2}-15\sqrt{2}+16\sqrt{2}=5\sqrt{2}\)
\(\sqrt{45}+\sqrt{20}-\sqrt{5}=3\sqrt{5}+2\sqrt{5}-\sqrt{5}=4\sqrt{5}\)
\(\sqrt{8}-\sqrt{32}+\sqrt{50}=2\sqrt{2}-4\sqrt{2}+5\sqrt{2=3\sqrt{2}}\)
\(\sqrt{12}+\sqrt{27}-4\sqrt{3}=2\sqrt{3}+3\sqrt{3}-4\sqrt{3}=\sqrt{3}\)

\(\sqrt{18}-\sqrt{50}=\sqrt{2.9}-\sqrt{25.2}=3\sqrt{2}-5\sqrt{2}=-2\sqrt{2}\)
\(\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\)
\(\sqrt{45}-2\sqrt{20}=\sqrt{9.5}-2\sqrt{4.5}=3\sqrt{5}-4\sqrt{5}=-\sqrt{5}\)
\(\sqrt{27}-\sqrt{12}=\sqrt{9.3}-\sqrt{3.4}=3\sqrt{3}-2\sqrt{3}=\sqrt{3}\)

bài 3
a. \(3\sqrt{16}-2\sqrt{9}+\dfrac{\sqrt{8}}{\sqrt{2}} =3.4-2.3+2=8\)
b.\(\sqrt{64}+4\sqrt{4}+2000=8+8+2000=2016\)

bài 2
a.
\(\dfrac{\sqrt{27}}{\sqrt{3}}=\sqrt{\dfrac{27}{3}}=\sqrt{9}=3\)
b.
\(\dfrac{\sqrt{25}}{\sqrt{121}}=\dfrac{5}{11}\)
c.
\(\sqrt{3}.\sqrt{27}=\sqrt{3.27}=\sqrt{81}=9\)
d.
\(\sqrt{25.49}=\sqrt{25}.\sqrt{49}=5.7=35\)

chỗ 12/100 k có căn