a. Mary is as tall as her brother.
b. A lemon is not sweet as an orange.
c. A donkey is not big as a horse.
d. This dress is as pretty as that one.
e. the weather is not cold as yesterday.
g. Mary does not sing as well as he.
h. They live as conveniently as we.
III.
1. My school is bigger than your school.
2. Lan is younger than Hoa.
3. My father is older than my mother.
4. This ruler is longer than that ruler.
5. This room is larger than my room.
6. The boys are stronger than the girls.
7. Linda cooks more terribly than her mother.
8. I do exercise more carefully than my brother.
9. He runs slowly than she. 

 II :

Mary is as tall as her brother.

A lemon is not as sweet as an orange.

A donkey is not as big as a horse.

This dress is as pretty as that one.

The weather is not as cold as yesterday.

  Mary does not sing as well as he.

 They live as conveniently as we.

Bài 5 hình 1: (tự vẽ hình nhé bạn)
a) Xét ΔABD và ΔACB ta có:
\(\widehat{BAD}\)= \(\widehat{BAC}\) (góc chung)
\(\widehat{ABD}\)= \(\widehat{ACB}\) (gt)
=> ΔABD ~ ΔACB (g-g)
=> \(\dfrac{AB}{AC}\) = \(\dfrac{BD}{CB}\) = \(\dfrac{AD}{AB}\) (tsđd)
b) Ta có: \(\dfrac{AB}{AC}\) = \(\dfrac{AD}{AB}\) (cm a)
=> \(AB^2\) = AD.AC
=> \(2^2\) = AD.4
=> AD = 1 (cm)
Ta có: AC = AD + DC (D thuộc AC)
      => 4   =   1   + DC
      => DC = 3 (cm)
c) Xét ΔABH và ΔADE ta có: 
   \(\widehat{AHB}\) = \(\widehat{AED}\) (=\(90^0\))
   \(\widehat{ADB}\) = \(\widehat{ABH}\) (ΔABD ~ ΔACB)
=> ΔABH ~ ΔADE
=> \(\dfrac{AB}{AD}\) = \(\dfrac{AH}{AE}\) = \(\dfrac{BH}{DE}\) (tsdd)
Ta có: \(\dfrac{S_{ABH}}{S_{ADE}}\) = \(\left(\dfrac{AB}{AD}\right)^2\)= \(\left(\dfrac{2}{1}\right)^2\)= 4
=> đpcm

Tiếp bài 5 hình 2 (tự vẽ hình)
a) Xét ΔABC vuông tại A ta có:
\(BC^2\) = \(AB^2\) + \(AC^2\)
\(BC^2\) = \(21^2\) + \(28^2\)
BC = 35 (cm)
b) Xét ΔABC và ΔHBA ta có:
\(\widehat{BAC}\) = \(\widehat{AHB}\) ( =\(90^0\))
\(\widehat{ABC}\) = \(\widehat{ABH}\) (góc chung)
=> ΔABC ~ ΔHBA (g-g)
=> \(\dfrac{AB}{BH}\) = \(\dfrac{BC}{AB}\) (tsdd)
=> \(AB^2\) = BH.BC
=> \(21^2\) = 35.BH
=> BH = 12,6 (cm)
c) Xét ΔABC ta có:
BD là đường p/g (gt)
=> \(\dfrac{AD}{DC}\) = \(\dfrac{AB}{BC}\) (t/c đường p/g)
Xét ΔABH ta có: 
BE là đường p/g (gt)
=> \(\dfrac{HE}{AE}\) = \(\dfrac{BH}{AB}\) (t/c đường p/g)
Mà: \(\dfrac{AB}{BC}\) = \(\dfrac{BH}{AB}\) (cm b)
=> đpcm
d) Ta có: \(\left\{{}\begin{matrix}\widehat{HBE}+\widehat{BEH}=90^0\\\widehat{ABD}+\widehat{ADB=90^0}\\\widehat{HBE}=\widehat{ABD}\end{matrix}\right.\)
=> \(\widehat{BEH}=\widehat{ADB}\)
Mà \(\widehat{BEH}=\widehat{AED}\) (2 góc dd)
Nên \(\widehat{ADB}=\widehat{AED}\)
=> đpcm

22/ \(\omega A=8\pi\)

\(A^2=x^2+\dfrac{v^2}{\omega^2}\Leftrightarrow A^2=3,2^2+\dfrac{\left(4,8\pi\right)^2}{\omega^2}\)

\(\Leftrightarrow\omega^2A^2=3,2^2\omega^2+23,04\pi^2\Leftrightarrow64\pi^2=3,2^2.\omega^2+23,04\pi^2\Leftrightarrow\omega=2\pi\left(rad/s\right)\)

\(\Rightarrow f=\dfrac{\omega}{2\pi}=\dfrac{2\pi}{2\pi}=1\left(Hz\right)\Rightarrow D.1Hz\)

23/ \(\omega A=20;\omega^2A=80\Rightarrow\left\{{}\begin{matrix}\omega=4\left(rad/s\right)\\A=5cm\end{matrix}\right.\)

\(\Rightarrow v=\omega\sqrt{A^2-x^2}=4.\sqrt{5^2-4^2}=12\left(cm/s\right)\Rightarrow A.12cm/s\)

Lần sau bạn chú ý viết đầy đủ đề.

1.

\(\sqrt{9+4\sqrt{5}-\sqrt{9-4\sqrt{5}}}=\sqrt{9+4\sqrt{5}-\sqrt{5-2\sqrt{4.5}+4}}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{(\sqrt{5}-\sqrt{4})^2}}=\sqrt{9+4\sqrt{5}-(\sqrt{5}-\sqrt{4})}\)

\(=\sqrt{9+4\sqrt{5}-\sqrt{5}+2}=\sqrt{11+3\sqrt{5}}\)

2.

\(\sqrt{8-2\sqrt{7}-\sqrt{8+2\sqrt{7}}}=\sqrt{8-2\sqrt{7}-\sqrt{7+2\sqrt{7}+1}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{(\sqrt{7}+1)^2}}\)

\(=\sqrt{8-2\sqrt{7}-\sqrt{7}-1}=\sqrt{7-3\sqrt{7}}\)

Phạm Mạnh Kiên: sửa lại theo ý bạn thì làm như sau:

1.

\(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}=\sqrt{5+2\sqrt{5}.\sqrt{4}+4}-\sqrt{5-2\sqrt{5}.\sqrt{4}+4}\)

\(=\sqrt{(\sqrt{5}+\sqrt{4})^2}-\sqrt{(\sqrt{5}-\sqrt{4})^2}=|\sqrt{5}+2|-|\sqrt{5}-2|\)

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

2.

\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}=|\sqrt{7}-1|-|\sqrt{7}+1|\)

\(=-2\)