\(a,\dfrac{x}{3x+6}=\dfrac{x}{3\left(x+2\right)}=\dfrac{x\left(x+2\right)}{3\left(x+2\right)^2}\\ \dfrac{5}{x^2+4x+4}=\dfrac{5}{\left(x+2\right)^2}=\dfrac{15}{3\left(x+2\right)^2}\\ b,\dfrac{5}{x^2-y^2+2x+1}=\dfrac{5}{\left(x-y+1\right)\left(x+y+1\right)}=\dfrac{5x}{x\left(x-y+1\right)\left(x+y+1\right)}\\ \dfrac{6}{x\left(x+y+1\right)}=\dfrac{6\left(x-y+1\right)}{x\left(x-y+1\right)\left(x+y+1\right)}\)

\(c,\dfrac{7x}{x^4-1}=\dfrac{7x}{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)}=\dfrac{7x\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)}\\ \dfrac{5x}{x^4+2x^2+1}=\dfrac{5x}{\left(x^2+1\right)^2}=\dfrac{5x\left(x-1\right)\left(x+1\right)}{\left(x^2+1\right)^2\left(x-1\right)\left(x+1\right)}\)

c: Ta có: \(\left(\dfrac{1}{2}\right)^{2x+1}=\dfrac{1}{8}\)

\(\Leftrightarrow2x+1=3\)

\(\Leftrightarrow2x=2\)

hay x=1

d: Ta có: \(\left(-\dfrac{1}{3}\right)^{x+3}=\dfrac{1}{81}\)

\(\Leftrightarrow x+3=4\)

hay x=1

1 A

2 D

3 A

4 A

5 C

6 B

7 C

8 C

2:

a: =>-3/4-x=-1/2

=>x=-3/4+1/2=-3/4+2/4=-1/4

b: =>x^2=36

=>x=6 hoặc x=-6

c: =>x-9/20=-5/12

=>x=-5/12+9/20=1/30

d: x+-3/4=-11/3

=>x=-11/3+3/4=-35/12

e: =>1/3:x=4/3+5/3=3

=>x=1/9

2A:

Xét ΔAHD vuông tại H và ΔCKB vuông tại K có

AD=CB

góc ADH=góc CBK

=>ΔAHD=ΔCKB

=>AH=CK

AH vuông góc BD

CK vuông góc BD

=>AH//CK

Xét tứ giác AHCK có

AH//CK

AH=CK

=>AHCK là hình bình hành

Ex 1:

1. listen

2. rains

3. will recognise

4. don’t want

5. does/ begin

6. will not

7. does/ watches

8. plant

9. play

10. leaves

Ex 2:

1. watching

2. listening

3. to buy

4. to speak

5. making

6. to eat

7. working

8. to call

9. to build

10. doing

ta có 33/131 < 33/217< 53/217

=> 33/131< 53/217

# Linh 2k7#

trả lời 

33/131>33/132=1/4

53/217<53/212=1/4 hay 53/217<1/4<33/131

Vậy 53/217<33/131

hc tốt ~:B~ 

theo đề bài

\(\frac{53}{217}\)<\(\frac{33}{131}\)

chúc bn 

hok tốt

Ta có: \(\left(x-3.5\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{10}\right)^4\ge0\forall y\)

Do đó: \(\left(x-3.5\right)^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(\dfrac{7}{2};\dfrac{1}{10}\right)\)

do 

\(\left(x-3.5\right)^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)

mà ta có \(\left(x-3.5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)

nên \(\left(x-3.5\right)^2+\left(y-\dfrac{1}{10}\right)^4=0\)

suy ra \(\left\{{}\begin{matrix}x-3,5=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}\end{matrix}\right.\)

tick mik nha

\(A=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot9^2}=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^6}=\dfrac{3}{4}\)

\(C=27\cdot\left(-\dfrac{3}{2}\right)^{-5}\cdot\left(-\dfrac{2}{5}\right)^{-4}:\left(\dfrac{2}{125}\right)^{-1}\)

\(=27\cdot\dfrac{-32}{243}\cdot\dfrac{625}{16}\cdot\dfrac{2}{125}\)

\(=\dfrac{-32}{9}\cdot\dfrac{1}{8}\cdot5\)

\(=-\dfrac{20}{9}\)