b) Ta có: \(1-\dfrac{530}{570}=\dfrac{40}{570}\)

\(1-\dfrac{531}{571}=\dfrac{40}{571}\)

mà \(\dfrac{40}{570}>\dfrac{40}{571}\)

nên \(1-\dfrac{530}{570}>1-\dfrac{531}{571}\)

\(\Leftrightarrow\dfrac{530}{570}< \dfrac{531}{571}\)

Tương tự như bài trên:

\(\dfrac{33}{131}=1-\dfrac{98}{131}< 1-\dfrac{98}{151}=\dfrac{53}{151}\)

a, vì 33<53

       131<151

=>\(\dfrac{33}{131}< \dfrac{53}{151}\)

b, Vì 530<531

       570<571

\(\Rightarrow\dfrac{530}{570}< \dfrac{531}{571}\)

55 mũ 66 < 66 mũ 55

55⁶⁶ >  66⁵⁵

mk gthik chắc bn k hiểu nên tốt nhất mk k gthik 

Ta có: 55^66 = (55^66)^11 =  ((11.5)^6)^11 = (11^6.5^6)^11 = (11^5.11.5^6)^11

           66^55 = (66^5)^11 = ((11.6)^5^11 = (11^5.6^5)^11

Vì 11.5^6 > 6^5 nên 55^66 > 66^55

Nếu đúng cho mk một like nha !

c) Để A>-1 thì A+1>0

\(\Leftrightarrow\dfrac{1-x}{x+1}+1>0\)

\(\Leftrightarrow\dfrac{1-x+x+1}{x+1}>0\)

\(\Leftrightarrow\dfrac{2}{x+1}>0\)

mà 2>0

nên x+1>0

hay x>-1

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>-1\\x\ne1\end{matrix}\right.\)

a) Ta có: \(A=\left(\dfrac{x+1}{x-1}-\dfrac{1-x}{x+1}+\dfrac{4x^2}{1-x^2}\right):\dfrac{2x^2-2}{x^2-2x+1}\)

\(=\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{4x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2\left(x^2-1\right)}{\left(x-1\right)^2}\)

\(=\dfrac{x^2+2x+1+x^2-2x+1-4x^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)

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

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

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

\(=\dfrac{1-x}{x+1}\)

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Để A=2 thì \(\dfrac{1-x}{x+1}=2\)

\(\Leftrightarrow1-x=2\left(x+1\right)\)

\(\Leftrightarrow1-x-2x-2=0\)

\(\Leftrightarrow-3x=1\)

hay \(x=-\dfrac{1}{3}\left(thỏa\right)\)

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

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

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