Bài 2:

\(a,\dfrac{-1}{3}.\dfrac{5}{7}=\dfrac{-5}{21}\\ b,\dfrac{1}{2}.\dfrac{-3}{4}=\dfrac{-3}{8}\\ c,\dfrac{19}{7}.\dfrac{7}{15}=\dfrac{19.1}{1.15}=\dfrac{19}{15}\\ d,\dfrac{5}{11}:\left(-9\right)=\dfrac{5}{11}.\dfrac{-1}{9}=\dfrac{-5}{99}\\ e,-1:\dfrac{3}{5}=-1.\dfrac{5}{3}=-\dfrac{5}{3}\\ f,\dfrac{-2}{9}:\dfrac{2}{9}:\dfrac{-9}{14}=-\left(\dfrac{2}{9}:\dfrac{2}{9}\right).\dfrac{-14}{9}=-1.\dfrac{-14}{9}=\dfrac{14}{9}\\ k,\left(-\dfrac{2}{7}\right)^2=\left(-1\right)^2.\left(\dfrac{2}{7}\right)^2=1.\dfrac{2^2}{7^2}=\dfrac{4}{49}\\ l,\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^3=\left(\dfrac{1}{3}\right)^4=\dfrac{1^4}{3^4}=\dfrac{1}{243}\\ m,\left(-\dfrac{1}{2}\right)^2.\left(\dfrac{1}{2}\right)^3=\left(-1\right)^2.\left(\dfrac{1}{2}\right)^{2+3}=1.\left(\dfrac{1}{2}\right)^5=1.\dfrac{1^5}{2^5}=\dfrac{1}{32}\)

Bài 1:

\(a,\dfrac{1}{-8}+\dfrac{-5}{8}=\dfrac{-1}{8}+\dfrac{-5}{8}=\dfrac{-\left(1+5\right)}{8}=-\dfrac{6}{8}=\dfrac{-6:2}{8:2}=-\dfrac{3}{4}\\ b,\dfrac{1}{7}+\dfrac{-3}{7}=\dfrac{1-3}{7}=-\dfrac{2}{7}\\ c,\dfrac{-12}{35}+\dfrac{-7}{35}=\dfrac{-\left(12+7\right)}{35}=\dfrac{-19}{35}\\ d,\dfrac{1}{6}+\dfrac{2}{5}=\dfrac{1.5+2.6}{6.5}=\dfrac{5+12}{30}=\dfrac{17}{30}\\ e,\dfrac{3}{5}+\dfrac{-7}{4}=\dfrac{3.4-7.5}{5.4}=\dfrac{12-35}{20}=\dfrac{-23}{20}\\ g,-2-\left(-\dfrac{1}{5}\right)=-2+\dfrac{1}{5}=\dfrac{-2.5+1}{5}=\dfrac{-9}{5}\\ h,\dfrac{2}{3}-\left(-1\right)=\dfrac{2}{3}+1=\dfrac{5}{3}\\ i,4-\dfrac{2}{3}=\dfrac{4.3-2}{3}=\dfrac{12-2}{3}=\dfrac{10}{3}\\ j,\dfrac{3}{4}-2=\dfrac{3-2.4}{4}=\dfrac{-5}{4}\\ k,-1-\left(-\dfrac{2}{3}\right)=-1+\dfrac{2}{3}=\dfrac{-1.3+2}{3}=\dfrac{-3+2}{3}=-\dfrac{1}{3}\)

16 C

17 A

18 B

19 A

20 C

21 B

22 B

23 C

24 B

25 D

26 D

27 B

28 C

29 A

30 D

31 C

Ta có: \(7x+4=x-2m\left(1\right)\)

Thay \(x=6\) vào \(\left(1\right)\) ta có:

\(7.6+4=6-2m\)

\(\Rightarrow46=6-2m\)

\(\Rightarrow-2m=40\)

\(\Rightarrow m=-20\)

a: Xét ΔEBF và ΔECD có

\(\widehat{EBF}=\widehat{ECD}\)(hai góc so le trong, BF//CD)

\(\widehat{BEF}=\widehat{CED}\)(hai góc đối đỉnh)

Do đó: ΔEBF~ΔECD(2)

Xét ΔEBF và ΔDAF có

\(\widehat{F}\) chung

\(\widehat{EBF}=\widehat{DAF}\)(hai góc đồng vị, BE//AD)

Do đó: ΔEBF~ΔDAF(1)

Từ (1) và (2) suy ra ΔECD~ΔDAF

b: BE+CE=BC

=>BE+4=6

=>BE=2(cm)

Xét ΔFAD có BE//AD

nên \(\dfrac{FB}{FA}=\dfrac{EB}{AD}\)

=>\(\dfrac{FB}{BF+15}=\dfrac{2}{6}=\dfrac{1}{3}\)

=>\(3BF=BF+15\)

=>2BF=15

=>BF=7,5(cm)

AF=AB+BF=15+7,5=22,5(cm)

c: Ta có: ΔECD~ΔDAF

=>\(\dfrac{EC}{DA}=\dfrac{DE}{DF}\)

=>\(EC\cdot DF=DE\cdot DA\)

Ta có: ΔECD~ΔDAF

=>\(\dfrac{CD}{AF}=\dfrac{EC}{DA}\)

=>\(EC\cdot AF=CD\cdot DA\)

1 I'd rather you didn't go out this Christmas

2 I'd rather the children went to bed

3 I'd rather you helped your mother with housework

4 I'd rather you didn't come to class late

5 I'd rather you kept silent in the classroom

It's time

1 It's time for you to dress yourself

2 It's about time the children went to bed

3 It's time we hurried up or we will be late for the train

4 It's high time children went to bed

5 It's about time for us to go home

C

\(SO_2+H_2O⇌H_2SO_3\)

\(K_2O+H_2O\rightarrow2KOH\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

a: 20inch=50,8cm=50,80cm

b: 30inch=76,20cm

where my father worked then

the good performance, she lost the match

, who sings Western folk songs very well, can compost songs

the first time I've ever read such an interesting book

where my father worked then

the good performance, she lost the match

, who sings Western folk songs very well, can compost songs

the first time I've ever read such an interesting book