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

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 \(AH\cdot BC=63\Rightarrow AH=\frac{63}{\sqrt{130}}\)

đẹp thế còn gì nữa. 

cảm ơn

Áp dụng định lý Pi-ta-go cho  \(\Delta ABH\)vuông tại H ta có : 

\(BH^2=AB^2-AH^2\)

\(\Leftrightarrow BH^2=13^2-5^2\)

\(\Leftrightarrow BH^2=144\)

\(\Leftrightarrow BH=12\)

Áp dụng hệ thức lượng trong tam giác ta có :

\(AB^2=BC.BH\)

\(\Leftrightarrow13^2=BC.12\)

\(\Leftrightarrow BC=\frac{169}{12}\)

Áp dụng định lí Py-ta-go cho  \(\Delta ABC\)vuông tại A ta có :

\(AC^2=BC^2-AB^2\)

\(\Leftrightarrow AC^2=\left(\frac{169}{12}\right)^2-13^2\)

\(\Leftrightarrow AC^2=\frac{4225}{144}\)

\(\Leftrightarrow AC=\frac{65}{12}\)

Ta có :  \(BH+CH=BC\)

\(\Leftrightarrow CH=BC-BH=\frac{169}{12}-12=\frac{25}{12}\)

Vậy  \(BC=\frac{169}{12};BH=12;AC=\frac{65}{12};CH=\frac{25}{12}\)

cảm ơn

Ta có: \(\dfrac{AB}{BC}=\dfrac{3}{5}\)

nên \(AB=\dfrac{3}{5}BC\)

Ta có: \(AB^2=BH\cdot BC\)

\(\Leftrightarrow\dfrac{9}{25}BC^2-a\cdot BC=0\)

\(\Leftrightarrow BC\cdot\left(\dfrac{9}{25}BC-a\right)=0\)

\(\Leftrightarrow BC\cdot\dfrac{9}{25}=a\)

hay \(BC=a:\dfrac{9}{25}=\dfrac{25}{9}a\)

\(\Leftrightarrow AB=\dfrac{3}{5}BC=\dfrac{3}{5}\cdot\dfrac{25}{9}a=\dfrac{5}{3}a\)

\(\Leftrightarrow CH=BC-BH=\dfrac{25}{9}a-a=\dfrac{16}{9}a\)

\(\Leftrightarrow AC=\sqrt{\left(\dfrac{25}{9}a\right)^2-\left(\dfrac{5}{3}a\right)^2}=\dfrac{20}{9}a\)

\(\Leftrightarrow AH=\sqrt{\left(\dfrac{20}{9}a\right)^2-\left(\dfrac{16}{9}a\right)^2}=\dfrac{4}{3}a\)

3: 

\(BC=\sqrt{12^2+16^2}=20\left(cm\right)\)

HB=12^2/20=7,2cm

=>HC=20-7,2=12,8cm

\(AD=\dfrac{2\cdot12\cdot16}{12+16}\cdot cos45=\dfrac{48\sqrt{2}}{7}\)

\(HD=\sqrt{AD^2-AH^2}=\dfrac{48}{35}\left(cm\right)\)