1 Your sister is interested in doing DIY in her free time, isn't she?

2 Unless that girl study harder, she will fail the exam

3 My father has given up smoking since last year

4 Mary wanted to know what Peter would give his brother on his birthday the day after

1. Your sister is interested in doing DIY in her free time, isn't she?

2. Unless that girl studies harder, she'll fail the exam.

3. My father has given up smoking since last year.

4. Mary wanted to know what Peter would give his brother on his birthday the next day.

1 F

2 T

3 T

4 T

5 F

6 F

1 F

2 T

3 T

4 T

5 F

6 F

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

Bài 1

a) \(\dfrac{3}{8}+\dfrac{5}{8}=\dfrac{8}{8}=1\)             

b) \(\dfrac{3}{4}+\dfrac{-7}{16}=\dfrac{12}{16}+\dfrac{-7}{16}=\dfrac{5}{16}\)

c) \(2\dfrac{17}{20}-\dfrac{1}{2}+3\dfrac{3}{20}=\dfrac{57}{20}-\dfrac{1}{2}+\dfrac{63}{20}\)\(=\dfrac{47}{20}+\dfrac{63}{20}=\dfrac{110}{20}=\dfrac{11}{2}\)

d) \(\dfrac{2}{3}-2\dfrac{1}{8}+\dfrac{7}{24}=\dfrac{2}{3}-\dfrac{17}{8}+\dfrac{7}{24}=\dfrac{16}{24}-\dfrac{51}{24}+\dfrac{7}{24}=\dfrac{16-51+7}{24}=\dfrac{-28}{24}=\dfrac{-7}{6}\)

Bài 2 :

a) \(x-\dfrac{7}{4}=3\)

\(x=3+\dfrac{7}{4}\)

\(x=\dfrac{19}{4}\)

b) \(x-\dfrac{1}{2}=\dfrac{4}{16}\cdot\dfrac{8}{3}\)

   \(x-\dfrac{1}{2}=\dfrac{2}{3}\)

  \(x=\dfrac{2}{3}+\dfrac{1}{2}\)

 \(x=\dfrac{5}{6}\)

c) \(\dfrac{15}{11}\div x=\dfrac{45}{22}\)

             \(x=\dfrac{15}{11}\div\dfrac{45}{22}\)

            \(x=\dfrac{2}{3}\)

d) \(\dfrac{8}{3}-2x=\dfrac{8}{5}-1\)

    \(\dfrac{8}{3}-2x=\dfrac{3}{5}\)

           \(2x=\dfrac{8}{3}-\dfrac{3}{5}\)

           \(2x=\dfrac{31}{15}\)

            \(x=\dfrac{31}{15}\div2\)

           \(x=\dfrac{31}{30}\)

Câu 4:

1: ΔOCD cân tại O

mà OI là đường trung tuyến

nên OI⊥CD tại I

Ta có: \(\hat{OIM}=\hat{OAM}=\hat{OBM}=90^0\)

=>O,I,A,M,B cùng thuộc đường tròn đường kính OM

2: Xét (O) có

MA,MB là các tiếp tuyến

Do đó: OM là phân giác của góc AOB

=>\(\hat{AOM}=\hat{BOM}\)

mà \(\hat{AOM}=\hat{AIM};\hat{BOM}=\hat{BIM}\)

nên \(\hat{AIM}=\hat{BIM}\)

=>IM là phân giác của góc AIB

sossssss

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