Xét ΔAEB vuông tại E và ΔAFC vuông tại F có

\(\hat{EAB}\) chung

Do đó: ΔAEB~ΔAFC

=>\(\frac{AE}{AF}=\frac{AB}{AC}\)

=>\(AE\cdot AC=AF\cdot AB\left(1\right)\)

Xét ΔAKC vuông tại K có KE là đường cao

nên \(AE\cdot AC=AK^2\left(2\right)\)

Xét ΔALB vuông tại L có LF là đường cao

nên \(AF\cdot AB=AL^2\left(3\right)\)

Từ (1),(2),(3) suy ra \(AK^2=AL^2\)

=>AK=AL

=>ΔALK cân tại A

=>\(\hat{AKL}=\hat{ALK}\)

Ta có: \(AL^2=AF\cdot AB\)

AL=AK

DO đó: \(AK^2=AF\cdot AB\)

=>\(\frac{AK}{AF}=\frac{AB}{AK}\)

Xét ΔAKB và ΔAFK có

\(\frac{AK}{AF}=\frac{AB}{AK}\)

góc KAB chung

Do đó: ΔAKB~ΔAFK

=>\(\hat{ABK}=\hat{AKF}\)

=>\(\hat{AKF}=\hat{ABE}\) (4)

Ta có: \(AK^2=AE\cdot AC\)

AK=AL

Do đó: \(AL^2=AE\cdot AC\)

=>\(\frac{AL}{AE}=\frac{AC}{AL}\)

Xét ΔALC và ΔAEL có

\(\frac{AL}{AE}=\frac{AC}{AL}\)

góc LAC chung

Do đó: ΔALC~ΔAEL

=>\(\hat{ACL}=\hat{ALE}\)

=>\(\hat{ALE}=\hat{ACF}\)

mà \(\hat{ACF}=\hat{ABE}\left(=90^0-\hat{BAC}\right)\)

nên \(\hat{ALE}=\hat{ABE}\) (5)

Từ (4),(5) suy ra \(\hat{ALE}=\hat{AKF}\)

mà \(\hat{ALK}=\hat{AKL}\)

nên \(\hat{ALE}+\hat{ALK}=\hat{AKL}+\hat{AKF}\)

=>\(\hat{ELK}=\hat{FKL}\)

Xét ΔAEB vuông tại E và ΔAFC vuông tại F có

\(\hat{EAB}\) chung

Do đó: ΔAEB~ΔAFC

=>\(\frac{AE}{AF}=\frac{AB}{AC}\)

=>\(AE\cdot AC=AF\cdot AB\left(1\right)\)

Xét ΔAKC vuông tại K có KE là đường cao

nên \(AE\cdot AC=AK^2\left(2\right)\)

Xét ΔALB vuông tại L có LF là đường cao

nên \(AF\cdot AB=AL^2\left(3\right)\)

Từ (1),(2),(3) suy ra \(AK^2=AL^2\)

=>AK=AL

=>ΔALK cân tại A

=>\(\hat{AKL}=\hat{ALK}\)

Ta có: \(AL^2=AF\cdot AB\)

AL=AK

DO đó: \(AK^2=AF\cdot AB\)

=>\(\frac{AK}{AF}=\frac{AB}{AK}\)

Xét ΔAKB và ΔAFK có

\(\frac{AK}{AF}=\frac{AB}{AK}\)

góc KAB chung

Do đó: ΔAKB~ΔAFK

=>\(\hat{ABK}=\hat{AKF}\)

=>\(\hat{AKF}=\hat{ABE}\) (4)

Ta có: \(AK^2=AE\cdot AC\)

AK=AL

Do đó: \(AL^2=AE\cdot AC\)

=>\(\frac{AL}{AE}=\frac{AC}{AL}\)

Xét ΔALC và ΔAEL có

\(\frac{AL}{AE}=\frac{AC}{AL}\)

góc LAC chung

Do đó: ΔALC~ΔAEL

=>\(\hat{ACL}=\hat{ALE}\)

=>\(\hat{ALE}=\hat{ACF}\)

mà \(\hat{ACF}=\hat{ABE}\left(=90^0-\hat{BAC}\right)\)

nên \(\hat{ALE}=\hat{ABE}\) (5)

Từ (4),(5) suy ra \(\hat{ALE}=\hat{AKF}\)

mà \(\hat{ALK}=\hat{AKL}\)

nên \(\hat{ALE}+\hat{ALK}=\hat{AKL}+\hat{AKF}\)

=>\(\hat{ELK}=\hat{FKL}\)

1. Our next -door neighbor _________ his car every Sunday.
A. is washing В. washes С. has washed D. is wash
2. Last summer, I _________ to the beach almost every day.
A. went В. was going С. have been D. have been going
3. 'Whose is this plane ticket on the floor?' 'Oh, it _________ to me. Thank you.'
A. is belonging В. belongs С. has belonged D. belonged
4. 'I'm really tired of travelling so much.' – 'I thought you ________ a bit quiet.'
A. were seeming В. have seemed
С. have been seeming D. seemed
5. 'You look thoughtful' – 'I _________ about our holiday last year.'
A. just think В. had just thought
С. am just think D. was just thinking
6. 'You went to Chile, didn't you?' – 'No, but I _________ to Peru, which is right next
door.'
A. had gone В. was gone С. did go D. was going
7. 'Why were you so tired yesterday?' 'Because I....... all morning.'
A. jog C. had been jogging
B. was jogged D. had been jogged
8. It was the first time I....... a live match.
A. was ever seeing C. had ever seen
B. had ever been seeing D. was ever seen
9. You live in a huge house,don't you?' 'Yes,but we.......!'
A. didn't use to C. use not
B. wouldn't D. weren't used to
10. I.......the whole of War and Peace by the time I was seven years old.
A. was reading C. had read
B. had been reading D. had been read

11. This time next week, we ___________the chemistry exam.
A have finished C will have finished
B have been finishing D will have been finishing
12. When I grow up, ______________an inventor.
A I'm being C I will have been
B I'm going to be D I will be being
13. Shirley will..........her research for the next few weeks.
A have done C have been doing
B be doing(tham khảo c13) D have been done
14. Next year, Sam will..........patients at this hospital for twenty-five years.
A have been treating C be treating
B treat D be going to treat

1. Our next -door neighbor _________ his car every Sunday.
A. is washing В. washes С. has washed D. is wash
2. Last summer, I _________ to the beach almost every day.
A. went В. was going С. have been D. have been going
3. 'Whose is this plane ticket on the floor?' 'Oh, it _________ to me. Thank you.'
A. is belonging В. belongs С. has belonged D. belonged
4. 'I'm really tired of travelling so much.' – 'I thought you ________ a bit quiet.'
A. were seeming В. have seemed
С. have been seeming D. seemed
5. 'You look thoughtful' – 'I _________ about our holiday last year.'
A. just think В. had just thought
С. am just think D. was just thinking
6. 'You went to Chile, didn't you?' – 'No, but I _________ to Peru, which is right next
door.'
A. had gone В. was gone С. did go D. was going
7. 'Why were you so tired yesterday?' 'Because I....... all morning.'
A. jog C. had been jogging
B. was jogged D. had been jogged
8. It was the first time I....... a live match.
A. was ever seeing C. had ever seen
B. had ever been seeing D. was ever seen
9. You live in a huge house,don't you?' 'Yes,but we.......!'
A. didn't use to C. use not
B. wouldn't D. weren't used to
10. I.......the whole of War and Peace by the time I was seven years old.
A. was reading C. had read
B. had been reading D. had been read

11. This time next week, we ___________the chemistry exam.
A have finished C will have finished
B have been finishing D will have been finishing
12. When I grow up, ______________an inventor.
A I'm being C I will have been
B I'm going to be D I will be being
13. Shirley will..........her research for the next few weeks.
A have done C have been doing
B be doing(tham khảo c13) D have been done
14. Next year, Sam will..........patients at this hospital for twenty-five years.
A have been treating C be treating
B treat D be going to treat

a: Gọi O là trung điểm của BC

=>O là tâm đường tròn đường kính BC

Xét (O) có

ΔBEC nội tiếp

BC là đường kính

Do đó: ΔBEC vuông tại E

=>CE⊥AB tại E và \(\hat{BEC}=90^0\)

Xét (O) có

ΔBFC nội tiếp

BC là đường kính

Do đó: ΔBFC vuông tại F

=>BF⊥AC tại F và \(\hat{BFC}=90^0\)

b: Xét ΔABC có

BF,CE là các đường cao

BF cắt CE tại H

Do đó: H là trực tâm của ΔABC

=>AH⊥BC

c: Kẻ OK⊥EF tại K

ΔOEF cân tại O

mà OK là đường cao

nên K là trung điểm của EF

Ta có: OK⊥EF

BM⊥EF

CN⊥EF

DO đó: BM//CN//OK

Xét hình thang BMNC có

O là trung điểm của BC

OK//BM//CN

Do đó: K là trung điểm của MN

Ta có: KE+EM=KM

KF+FN=KN

mà KE=KF và KM=KN

nên EM=FN

Ta có:

\(AH^2=BH.HC\Rightarrow HC=\dfrac{AH^2}{BH}=\dfrac{3^2}{4}=\dfrac{9}{4}\left(cm\right)\)

\(BC=BH+HC=4+\dfrac{9}{4}=9\left(cm\right)\)

\(AB=\sqrt{BH.BC}=\sqrt{4.9}=6\left(cm\right)\)

\(AC=\sqrt{CH.BC}=\sqrt{\dfrac{9}{4}.9}=\dfrac{9}{2}\left(cm\right)\)

bạn gõ đề bằng latex để rõ đề

\(a,\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\) \(\left(dk:x\ge0,x\ne1\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\\ =\sqrt{x}\)

\(b,\dfrac{1-2\sqrt{x}+x}{1-\sqrt{x}}\left(dkxd:x\ge0,x\ne1\right)\)

\(=\dfrac{1^2-2\sqrt{x}+\sqrt{x^2}}{1-\sqrt{x}}\\ =\dfrac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}\\ =1\)