a/

\(x-y=\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}=\frac{1}{bd}.\) (1)

\(y-z=\frac{c}{d}-\frac{e}{h}=\frac{ch-de}{dh}=\frac{1}{dh}\)(2)

+ Nếu d>0 => (1)>0 và (2)>0 => x>y; y>x => x>y>z

+ Nếu d<0 => (1)<0 và (2)<0 => x<y; y<z => x<y<z

b/

\(m-y=\frac{a+e}{b+h}-\frac{c}{d}=\frac{ad+de-cb-ch}{d\left(b+h\right)}=\frac{\left(ad-cb\right)-\left(ch-de\right)}{d\left(b+h\right)}=\frac{1-1}{d\left(b+h\right)}=0\)

=> m=y

+

cảm ơn bn nha Nguyễn Ngoc Anh Minh mk k cho bn r đó kb vs mk nha

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a} = \frac{a+b+c}{b+c+a}=1\) (tính chất của dãy tỉ số bằng nhau)

=> a=b=c

chúc bn học giỏi

ta có 

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow a=b=c\)

TC: 

10A = \(\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1}{10^{12}-1}-\frac{9}{10^{12}-1}=1-\frac{9}{10^{12}-1}< 1\)

10B = \(\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1}{10^{11}+1}+\frac{9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)

VÌ  \(1-\frac{9}{10^{12}-1}< 1\)VÀ \(1+\frac{9}{10^{11}+1}>1\) nên \(1+\frac{9}{10^{11}+1}\)\(>\)\(1-\frac{9}{10^{12}-1}\)

\(=>\)\(10A< 10B\)

\(=>A< B\)

Vậy \(A< B\)

mai mình làm típ cho

\(A=\left|x-\frac{1}{3}\right|+\frac{1}{4}\ge\frac{1}{4}>\frac{1}{5}\)

khi a,b khác giấu và b#0 ta có 

-a/b hoặc a/-b

vì hai số hữu  tỉ là số âm nên=>a/b<0

\(A=\frac{17^{18}+1}{17^{19}+1}\)

\(17A=\frac{17^{19}+17}{17^{19}+1}=\frac{\left(17^{19}+1\right)+16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

\(B=\frac{17^{17}+1}{17^{18}+1}\)

\(17B=\frac{17^{18}+17}{17^{18}+1}=\frac{\left(17^{18}+1\right)+16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

\(\text{Vì}\)\(1+\frac{16}{17^{19}+1}< 1+\frac{16}{17^{18}+1}\)

\(\Leftrightarrow17A< 17B\)

\(\Leftrightarrow A< B\)

Trả lời

\(17A=\frac{\left(17^{18}+1\right)17}{17^{19}+1}=\frac{17^{19}+17}{17^{19}+1}=\frac{17^{19}+1+16}{17^{19}+1}=\frac{17^{19}+1}{17^{19}+1}+\frac{16}{17^{19}+1}=1+\frac{16}{17^{19}+1}\)

\(17B=\frac{\left(17^{17}+1\right)17}{17^{18}+1}=\frac{17^{18}+17}{17^{18}+1}=\frac{17^{18}+1+16}{17^{18}+1}=\frac{17^{18}+1}{17^{18}+1}+\frac{16}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)

Vì \(17^{19}+1>17^{18}+1\)

\(\Rightarrow\frac{16}{17^{18}+1}>\frac{16}{17^{19}+1}\)

\(\Rightarrow1+\frac{16}{17^{18}+1}>1+\frac{16}{17^{19}+1}\)

\(\Rightarrow B>A\)

cách 1

A=1/3=5/15

B=6/5=18/15

vi 18/15>5/15

=>B>A

cách 2

vi A <1

B>1

=>A<B

Ta có: A = 1/3 < 1

Ta có: B = 6/5 > 1

=> A < B

1/3 .......6/5

= mẫu số chung là 15 

= 1/3 x 5/5 

= 6/5 x 3/3

= 5/15 .......18/15

Vậy : 5/15 < 18/15

Ta co:

         B=\(\frac{10^{30}+1}{10^{31}+1}\)<\(\frac{10^{30}+1+99}{10^{31}+1+99}\)=\(\frac{10^{30}+100}{10^{31}+100}\)=\(\frac{10^{10}\cdot\left(10^{20}+1\right)}{10^{10}\cdot\left(10^{21}+1\right)}\)=\(\frac{10^{20}+1}{10^{21}+1}\)=A

Vay A<B