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27/12/2017 lúc 18:59
Ex1: Điền từ thích hợp vào chỗ trống
This is Ba. He(1)......... a student.Every morning he(2).........up at 5.30.He(3).............. his teeth and takes a(4)............... then has breakfast at 6.15. He goes to school(5)........six thirty.His house is(6).............his house so he walks.The classes(7)............at 7.15 and finish at 11.15.In the afternoon he plays sports with his friend,Nam. They play badminton but now they(8).................soccer.In the evening he (9)......his homework and goes to(10).........at 9.30
Ex2:Cho dạng đúng của động từ trong ngoặc
1.My sister(have)...........classes from Monday to Friday
2.She(read)................a book in her room now
3.He(get)........................up at 6.00 every day?
4.There(not be)..............a big yard behind his classroom
27/12/2017 lúc 18:59
Ex1: Điền từ thích hợp vào chỗ trống
This is Ba. He(1)......... a student.Every morning he(2).........up at 5.30.He(3).............. his teeth and takes a(4)............... then has breakfast at 6.15. He goes to school(5)........six thirty.His house is(6).............his house so he walks.The classes(7)............at 7.15 and finish at 11.15.In the afternoon he plays sports with his friend,Nam. They play badminton but now they(8).................soccer.In the evening he (9)......his homework and goes to(10).........at 9.30
Ex2:Cho dạng đúng của động từ trong ngoặc
1.My sister(have)...........classes from Monday to Friday
2.She(read)................a book in her room now
3.He(get)........................up at 6.00 every day?
4.There(not be)..............a big yard behind his classroom
27/12/2017 lúc 18:59
Ex1: Điền từ thích hợp vào chỗ trống
This is Ba. He(1)......... a student.Every morning he(2).........up at 5.30.He(3).............. his teeth and takes a(4)............... then has breakfast at 6.15. He goes to school(5)........six thirty.His house is(6).............his house so he walks.The classes(7)............at 7.15 and finish at 11.15.In the afternoon he plays sports with his friend,Nam. They play badminton but now they(8).................soccer.In the evening he (9)......his homework and goes to(10).........at 9.30
Ex2:Cho dạng đúng của động từ trong ngoặc
1.My sister(have)...........classes from Monday to Friday
2.She(read)................a book in her room now
3.He(get)........................up at 6.00 every day?
4.There(not be)..............a big yard behind his classroom
Dễ quá đi
![](https://rs.olm.vn/images/avt/0.png?1311)
a. - Áp dụng hệ thức về cạnh và đường cao trong Δ vuông vào ΔABC vuông tại A ta có :
\(AH=\sqrt{CH.BH}=\sqrt{2.4}=2\sqrt{2}\) ( Đ.lý 2 )
- Áp dụng đ.lý Pytago vào \(\Delta AHB\perp H\) ta có :
\(AB=\sqrt{AH^2+BH^2}=\sqrt{\left(2\sqrt{2}\right)^2+4^2}=2\sqrt{6}\)
- \(BC=2+4=6\)
- Theo đ.lý Pytago :
\(AC=\sqrt{BC^2-AB^2}=\sqrt{6^2-\left(2\sqrt{6}\right)^2}=2\sqrt{3}\)
b. - Áp dụng hệ thức...trong Δ vuông ABC ta có :
+ \(BC=\dfrac{AB^2}{BH}=\dfrac{12^2}{6}=24\) ( Đ.lý 1 )
\(\Rightarrow CH=BC-BH=24-6=18\)
+ \(AH=\sqrt{BH.CH}=\sqrt{6.18}=6\sqrt{3}\) ( Đ.'ý 2 )
- Theo đ.lý Pytago ta có :
\(AC=\sqrt{BC^2-AB^2}=\sqrt{24^2-12^2}=12\sqrt{3}\)
a, BC = BH+HC
*\(AB=\sqrt{BH.BC}=\sqrt{4.8}=\sqrt{32}\)
*\(AC=\sqrt{HC.BC}=\sqrt{2.8}=4\)
*\(AH=\sqrt{BH.HC}=\sqrt{4.2}=\sqrt{8}\)
b,Theo định lý pytago ta có:
\(AH=\sqrt{AB^2-BH^2}=\sqrt{12^2-6^2}=6\sqrt{3}\)
*\(BC=\dfrac{AB^2}{BH}=\dfrac{12^2}{6}=2\)
*\(CH=BC-BH=24-6=18\)
\(AC=\sqrt{HC.BC}=\sqrt{18.24}=12\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Áp dụng định lí Pytago trong \(\Delta\) AHC vuông tại H ta có :
\(AH^2+HC^2=AC^2\)
\(\Rightarrow HC^2=AC^2-AH^2\)
\(\Rightarrow HC=\sqrt{AC^2-AH^2}=\sqrt{40^2-24^2}=32cm\)
b) Áp dụng định lí Pytago trong \(\Delta\) AHC vuông tại H ta có :
\(AH^2+HC^2=AC^2\)
\(\Rightarrow AC=\sqrt{AH^2+HC^2}=\sqrt{9,6^2+12,8^2}=16cm\)
c) \(BC=CH+BH=72+12,5=84,5\left(cm\right)\)
Ta có: \(\left\{{}\begin{matrix}AB^2=BH.BC=12,5.84,5=1056,25\\AC^2=CH.BC=72.84,5=6084\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}AB=\dfrac{65}{2}\left(cm\right)\\AC=78\left(cm\right)\end{matrix}\right.\)
Ta có: \(AB.AC=AH.BC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{78.\dfrac{65}{2}}{84,5}=30\left(cm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
VẼ HÌNH HƠI XẤU THÔNG CẢM NHA
áp dụng hệ thức lượng trong tam giác vuông ABC ta có \(AB\cdot AC=AH\cdot BC\) \(\Rightarrow AH\cdot BC=63\) (1)
áp dụng đl pitagovao tam giác vuông ABC ta có \(AB^2+AC^2=BC^2\Rightarrow BC=\sqrt{130}\)
thay vao (1) ta co \(AH\cdot BC=63\Rightarrow AH=\frac{63}{\sqrt{130}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, AB = 7,5cm, AC = 10cm, BC = 12,5cm, HC = 8cm
b, AH = 3 3 cm; P A B C = 18 + 6 3 c m ; P A B H = 9 + 3 3 c m ; P A C H = 9 + 9 3 c m
Xét tam giác ABC vuông tại A, đường cao AH
* Áp dụng hệ thức :
\(\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}\)mà \(\frac{AB}{AC}=\frac{3}{4}\Rightarrow AB=\frac{3}{4}AC\)( gt )
\(\Rightarrow\frac{1}{36}=\frac{1}{\left(\frac{3}{4}AC\right)^2}+\frac{1}{AC^2}\)
\(\Leftrightarrow\frac{1}{36}=\frac{AC^2+\left(\frac{3}{4}AC\right)^2}{AC^2\left(\frac{3}{4}AC\right)^2}\Rightarrow36AC^2+36\left(\frac{3}{4}AC\right)^2=AC^2\left(\frac{3}{4}AC\right)^2\)
\(\Leftrightarrow36AC^2+\frac{81}{4}AC^2=\frac{9}{16}AC^4\)
\(\Leftrightarrow\frac{225}{4}AC^2=\frac{9}{16}AC^4\Leftrightarrow\frac{9}{16}AC^4-\frac{225}{4}AC^2=0\)
\(\Leftrightarrow\frac{9}{16}AC^2-\frac{225}{4}=0\Leftrightarrow AC^2=\frac{225}{4}.\frac{16}{9}=25.4=100\Leftrightarrow AC=10\)cm
\(\Rightarrow AB=\frac{3}{4}AC\Rightarrow AB=\frac{3}{4}.10=\frac{30}{4}=\frac{15}{2}\)cm
* Áp dụng định lí Pytago ta có :
\(AB^2+AC^2=BC^2\Rightarrow BC^2=\frac{225}{4}+100=\frac{625}{4}\Rightarrow BC=\frac{25}{2}\)
* Áp dụng hệ thức : \(AB^2=BH.BC\Rightarrow BH=\frac{AB^2}{BC}=\frac{\frac{225}{4}}{\frac{25}{2}}=\frac{225}{4}.\frac{2}{25}=\frac{9}{2}\)
\(\Rightarrow CH=BC-BH=\frac{25}{2}-\frac{9}{2}=\frac{16}{2}=8\)
Vậy BH = 9/2 cm ; CH = 8 cm