A B C H D I

GT:AH vuông BC

      AD=AB

     DI vuông AH

KL:BH=ID

                                                    Bài làm

Ta có:

\(\widehat{A1}=\widehat{A2}\)(đối đỉnh)(1)

\(AB=AD\)(GT)(2)

mà\(\widehat{B}=180^0-90^0-\widehat{A1}\)

         \(\widehat{D}=180^0-90^0-\widehat{A2}\)

và\(\widehat{A1}=\widehat{A2}\)

=>\(\widehat{B}=\widehat{D}\)(3)

Từ (1),(2),(3) suy ra:\(\Delta\)ABH=\(\Delta\)ADI(g-c-g)

=>BH=ID(hai cạnh tương ứng)

                      Vậy BH=ID

Hk tot ^3^

Hình vẽ : tự vẽ nha 

Sabc = \(\frac{1}{2}AH.BC=\frac{1}{2}BK.AC\)

=> \(AH.BC=AC.BK\)

=> \(\frac{AH}{BK}=\frac{AC}{BC}=\frac{15.6}{12}=\frac{13}{10}\)

=> \(\frac{AC}{13}=\frac{BC}{10}=t\)

=> \(AC=13t;BC=10t\)

Tam giác ABC cân có AH là đg cao => AH là t tuyến => BH = HC = 1/2 BC = 1/2.10t = 5t 

TAm giác AHC vuông tại H , theo py ta go :

 \(AC^2-HC^2=15.6^2\)

=> \(169t^2-25t^2=15.6^2\)

tính ra t thay vào tìm ra BC 

2S_ABC = AH.BC = AC.BK

ð       15,6BC = 12AC

ð       BC = 12/15,6AC

ð       CH = 6/15,6AC

ð       AH² = AC² – HC² = 144/169AC²

ð       AH = 12/13 AC

ð       15,6 = 12/13AC

ð       AC = 16,9

ð       BC = 12/15,6 AC = 13

1 F

2 T

3 T

4 T

5 F

6 F

1 F

2 T

3 T

4 T

5 F

6 F

1 A

2 B

3 A

4 A

5 A

6 C

7 D

8 B

Bài 5:

a: Xét ΔABC và ΔCDA có

\(\hat{ACB}=\hat{DAC}\) (hai góc so le trong, AD//BC)

AC chung

\(\hat{CAB}=\hat{ACD}\) (hai góc so le trong, AB//CD)

Do đó: ΔABC=ΔCDA

b: ΔABC=ΔCDA

=>AB=CD; BC=AD

Xét ΔMAD và ΔMCB có

\(\hat{MAD}=\hat{MCB}\) (hai góc so le trong, AD//CB)

AD=CB

\(\hat{MDA}=\hat{MBC}\) (hai góc so le trong, AD//CB)

Do đó: ΔMAD=ΔMCB

=>MA=MC và MD=MB

c: Xét ΔMAE và ΔMCN có

MA=MC

\(\hat{MAE}=\hat{MCN}\) (hai góc so le trong, AE//CN)

AE=CN

Do đó: ΔMAE=ΔMCN

=>\(\hat{AME}=\hat{CMN}\)

mà \(\hat{AME}+\hat{EMC}=180^0\) (hai góc kề bù)

nên \(\hat{EMC}+\hat{CMN}=180^0\)

=>E,M,N thẳng hàng

d: DC⊥BC

DC//AB

Do đó: AB⊥BC

III

1 appreciated

2 worried

3 tense

4 confident

5 delighted

6 frustrated

7 calm

8 relaxed

9 self-disciplined

10 depressed

V

1 had resolved

2 has taken

3 Did you say

4 not to take

5 wanted

6 is being repaired

7 taking

8 to think

9 had worked

10 was washing - dropped

11 will find

12 faces

13 turned

14 to give

15 will empathise

Đặt \(A=\dfrac{1}{\sqrt[3]{a+7b}}+\dfrac{1}{\sqrt[3]{b+7c}}+\dfrac{1}{\sqrt[3]{c+7a}}\)

\(A=\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(a+7b\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(b+7c\right)}}+\dfrac{\sqrt[3]{64}}{\sqrt[3]{8.8\left(c+7a\right)}}\)

\(\ge\dfrac{4}{\dfrac{8+8+a+7b}{3}}+\dfrac{4}{\dfrac{8+8+b+7c}{3}}+\dfrac{4}{\dfrac{8+8+c+7a}{3}}\ge\dfrac{\left(2+2+2\right)^2}{\dfrac{8+8+a+7b+8+8+b+7c+8+8+c+7a}{3}}\)

\(=\dfrac{36.3}{8\left(a+b+c\right)+48}=\dfrac{3}{2}\)

Vậy \(A_{min}=\dfrac{3}{2}\Leftrightarrow a=b=c=1\)

I

1 D

2 C

3 A

II

4 C

5 B

Part B

6 A

7 D

8 B

9 C

10 A

11 C

12 C

13 B

14 C

15 D

16 A

17 B

18 D

19 C

20 A

1 A

2 C

3 D

4 C

5 D

6 D

7 A

8 D

9 B

10 A

11 B

12 A

13 A

14 C

15 C

16 A

17 A

18 B

19 A

20 D