Ta có A - 1 = \(\frac{10^5+4}{10^5-5}-1=\frac{10^5+4-10^5+5}{10^5-5}=\frac{9}{10^5-5}\)

Lại có : B -  1 = \(\frac{10^5+3}{10^5-6}-1=\frac{10^5+3-10^5+6}{10^5-6}=\frac{9}{10^5-6}\)

Vì \(\frac{9}{10^5-5}< \frac{9}{10^5-6}\Rightarrow A-1< B-1\Rightarrow A< B\)

Ta có : 

\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)

\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)

\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)

\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)

\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\) 

\(\Rightarrow\)\(S>10\) 

Vậy \(S>10\)

Chúc bạn học tốt ~ 

a) A=\(\frac{178}{179}+\frac{179}{180}+\frac{183}{181}\)

ta có :

 \(A=\left(1-\frac{1}{179}\right)+\left(1-\frac{1}{180}\right)+\left(1+\frac{2}{181}\right)\)

 \(\Rightarrow A=\left(1+1+1\right)-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)\)

\(\Rightarrow A=3-\left(\frac{1}{179}-\frac{1}{180}+\frac{2}{181}\right)< 3\)

Vậy \(A< 3\)

a. Ta có :

\(\frac{178}{179}< 1\left(\frac{1}{179}\right)\)

\(\frac{179}{180}< 1\left(\frac{1}{180}\right)\)

\(\frac{183}{181}>1\left(\frac{3}{181}\right)\left(1\right)\)

Mà \(\frac{3}{181}>\frac{1}{179}+\frac{1}{180}\left(=\frac{359}{32220}< \frac{3}{181}\right)\left(2\right)\)

Từ \(\left(1\right)\&\left(2\right)\Rightarrow\frac{178}{179}+\frac{179}{180}+\frac{183}{181}< 1+1+1\)

Vậy \(A< 3\)

b) \(A=\frac{1+5+5^2+5^3+...+5^{10}+5^{11}}{1+5+5^2+5^3+...+5^9+5^{10}}=5^{11}\)

bn rút gọn là dc  

\(B=\frac{1+7+7^2+7^3+...+7^{10}+7^{11}}{1+7+7^2+7^3+...+7^9+7^{10}}=7^{11}\)

\(A=5^{11},B=7^{11}\)

\(\Rightarrow7^{11}>5^{11}\Rightarrow B>A\)

hk tốt #

ta có :

ts của a=tử số của b

mà ms của a<ms của b

suy ra a>b

sai bét

ANH HUY

a=(10^7 -8 +13)/(10^7 - 8) = 1+ 13/(10^7 - 8)

b = (10^5 +6)/(10^5 -7) = (10^5-7+13)/(10^5 -7) = 1 + 13/(10^5-7)

vay b>a

Ta có:

1 = \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+............+\frac{1}{10}\)(10 phân số \(\frac{1}{10}\))

Mà \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};............;\frac{9}{10}>10\)

\(\Rightarrow M>1\)

Vậy M > 1

Ta có:

1/2=0,5

2/3>0,6

<=>1/2+2/3>1,1>1

<=>1/2+2/3+3/4+...+9/10>1

Vì 1 = \(\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)

\(\Rightarrow\)M > 1 vì \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};...;\frac{9}{10}>\frac{1}{10}\)

\(\Rightarrow M>1\)

(10^5+4)/(10^5-1)=(10^5-1+5)/(10^5-1)={(10^5-1)/(10^5-1)}+{5/(10^5-1)}=1+{5/(10^5-1)}    (1)

(10^5+3)/(10^5-2)=(10^5-2+5)/(10^5-2)={(10^5-2)/(10^5-2)}+{5/(10^5-2)}=1+{5/(10^5-2)}    (2)

từ 1 và 2 ta so sánh{5/(10^5-1)} và {5/(10^5-2)}....

suy ra ... kết quả

có thẻ 50k ko anh giải cho

nạp thẻ cho anh à

dễ thôi

A=\(\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)

B=\(\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)

\(10^8>10^7nen10^8-7>10^7-8\)

=> \(\frac{13}{10^8-7}< \frac{13}{10^7-8}hayB< A\)

\(\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1-\frac{13}{10^7-8}\);\(\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1-\frac{13}{10^7-7}\)

Vì \(\frac{13}{10^8-8}< \frac{13}{10^7-7}\)nên A>B

Ta có : 

\(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)

\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)

Mà \(\frac{13}{10^7-8}>\frac{13}{10^8-7}\left(10^7-8< 10^8-7\right)\)

\(\Rightarrow1+\frac{13}{10^7-8}>1+\frac{13}{10^8-7}\)

\(\Rightarrow A< B\)

Vậy \(A< B\)

~ Ủng hộ nhé