nhanh hộ mik đc ko ạ? Mik đang cần gấp 

a)2/15<8/17

b)12/17<13/18

c)9/20>3/7

d)17/18>15/19

@Trần Minh Tuấn Hữu co thể viết ra cách giải hộ mik đc ko?

a) \(\dfrac{12}{14}=\dfrac{1200}{1400}=\dfrac{1400-200}{1400}=1-\dfrac{200}{1400}\)

\(\dfrac{1212}{1414}=\dfrac{1414-200}{1414}=1-\dfrac{200}{1414}\)

vì \(\dfrac{200}{1414}< \dfrac{200}{1400}\)

Nên \(1-\dfrac{200}{1400}< 1-\dfrac{200}{1414}\)

Vậy \(\dfrac{12}{14}< \dfrac{1212}{1414}\)

Các bài sau tương tự

\(A=\dfrac{2}{\sqrt{17}+\sqrt{15}}\) ; \(B=\dfrac{2}{\sqrt{15}+\sqrt{13}}\)

Mà \(\sqrt{17}+\sqrt{15}>\sqrt{15}+\sqrt{13}>0\)

\(\Rightarrow\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{15}+\sqrt{13}}\)

\(\Rightarrow A< B\)

\(A=\sqrt{17}-\sqrt{15}=\dfrac{2}{\sqrt{17}+\sqrt{15}}\)

\(B=\sqrt{15}-\sqrt{13}=\dfrac{2}{\sqrt{13}+\sqrt{15}}\)

mà \(\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{13}+\sqrt{15}}\)

nên A<B

1) \(\frac{13}{17}< 1< \frac{17}{15}\)

2) Bài toán tổng tỉ: a:b=4 chọn a có 4 phần b có 1 phần

Tổng số phần : 4+1=5

a=4:5x4=16/5

b=4:5x1=4/5

1. Vì \(\frac{13}{17}< 1\) và \(\frac{17}{15}>1\)

nên \(\frac{13}{17}< 1< \frac{17}{15}\)

hay \(\frac{13}{17}< \frac{17}{15}\)

2. Ta có: \(a\div b=\frac{a}{b}=4\)

Gọi k = (a, b)

\(\Rightarrow\hept{\begin{cases}a=4k\\b=k\end{cases}}\)

Mà \(a+b=4\)

\(\Leftrightarrow4k+k=4\)

\(\Leftrightarrow5k=4\)

\(\Leftrightarrow k=\frac{4}{5}\)

\(\Rightarrow\hept{\begin{cases}a=4.\frac{4}{5}=\frac{16}{5}\\b=\frac{4}{5}\end{cases}}\)

A lớn hơn 4

B bé hơn 3

A = 15/14 + 16/15 + 17/16 + 18/17

Ta thấy :

15/14 > 1

16/15 > 1

17/16 > 1

18/17 > 1

=> A > 4

B tương tự

Ta có:

A =\(\frac{15}{14}< 4\)\(;\)\(\frac{16}{15}< 4\)\(;\)\(\frac{17}{16}< 4\)\(;\)\(\frac{18}{17}< 4\)

Vậy A bé hơn 4.

B =\(\frac{2015}{2016}< 3\)\(;\)\(\frac{2016}{2017}< 3\)\(;\)\(\frac{2016}{2019}< 3.\)

Vậy B bé hơn 3.

áp dụng tc \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{a+m}< 1\left(m\in N\right)\)

Ta có: \(B=\frac{15^{16}+1}{15^{17}+1}< \frac{15^{16}+1+14}{15^{17}+1+14}\)\(=\frac{15^{16}+15}{15^{17}+15}=\frac{15.\left(15^{15}+1\right)}{15.\left(15^{16}+1\right)}=\frac{15^{15}+1}{15^{16}+1}\)

\(\Rightarrow B< A\)

\(A=\frac{15^{15}+1}{15^{16}+1}\)

\(\Rightarrow15A=\frac{15^{16}+15}{15^{16}+1}\)

\(\Rightarrow15A=\frac{15^{16}+1+14}{15^{16}+1}\)

\(\Rightarrow15A=\frac{15^{16}+1}{15^{16}+1}+\frac{14}{15^{16}+1}\)

\(\Rightarrow15A=1+\frac{14}{15^{16}+1}\)

\(B=\frac{15^{16}+1}{15^{17}+1}\)

\(\Rightarrow15B=\frac{15^{17}+15}{15^{17}+1}\)

\(\Rightarrow15B=\frac{15^{17}+1+14}{15^{17}+1}\)

\(\Rightarrow15B=\frac{15^{17}+1}{15^{17}+1}+\frac{14}{15^{17}+1}\)

\(\Rightarrow15B=1+\frac{14}{15^{17}+1}\)

Vì \(\frac{14}{15^{17}+1}< \frac{14}{15^{16}+1}\) nên \(15B< 15A\)

Vậy B < A