Bài 6: So sánh phân số

ai trả lời đúng mình tick cho!!

Xét vế phải :

\(VT=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)

\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)

\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\) 

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

 \(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)

\(\text{Nhầm xíu , cho sửa lại nhé}\)

\(\text{Xét vế phải :}\)

\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)

\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)

\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)

\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)

\(A=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=2\cdot\dfrac{49}{100}=\dfrac{98}{100}>\dfrac{1}{4}\)

Ối trời !Sao mà dài thế này

Làm sao làm cho nổi

Help me! Please!

1: \(B=\dfrac{1}{7}-\dfrac{1}{9}=\dfrac{2}{63}>\dfrac{1}{7\cdot9}=A\)

2: \(A=\dfrac{15}{301}< \dfrac{15}{300}=\dfrac{1}{20}=\dfrac{25}{500}< \dfrac{25}{499}\)

a) ta có:

11/27 <11/16=33/48<33/47

=>11/27 <33/47

\(A=\dfrac{10}{a^m}+\dfrac{10}{a^n}\)

\(=\dfrac{10a^n+9a^m+a^m}{a^ma^n}\)

\(B=\dfrac{11}{a^m}+\dfrac{9}{a^n}\)

\(=\dfrac{10a^n+a^n+9a^m}{a^ma^n}\)

+ Nếu m > n thì a^m > aⁿ. \(\Rightarrow\) \(\dfrac{10a^n+9a^m+a^m}{a^ma^n}>\dfrac{10a^n+a^n+9a^m}{a^ma^n}\) hay A > B

+ Nếu m < n thì a^m < aⁿ. \(\Rightarrow\) \(\dfrac{10a^n+9a^m+a^m}{a^ma^n}< \dfrac{10a^n+a^n+9a^m}{a^ma^n}\) hay A < B

+ Nếu m = n thì a^m = aⁿ. \(\Rightarrow\) \(\dfrac{10a^n+9a^m+a^m}{a^ma^n}=\dfrac{10a^n+a^n+9a^m}{a^ma^n}\) hay A = B

\(A=\dfrac{8^9+12}{8^9+7}=\dfrac{8^9+7+5}{8^9+7}=\dfrac{8^9+7}{8^9+7}+\dfrac{5}{8^9+7}=1+\dfrac{5}{8^9+7}\left(1\right)\)

\(B=\dfrac{8^{10}+4}{8^{10}-1}=\dfrac{8^{10}-1+5}{8^{10}-1}=\dfrac{8^{10}-1}{8^{10}-1}+\dfrac{5}{8^{10}-1}=1+\dfrac{5}{8^{10}-1}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)

b: 51/59<1<513/293

c: 31/157<31/155=1/5

17/83>17/85=1/5

Do đó: 31/157<17/83