(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

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 #

\(A=\frac{10^{2011}+5}{10^{2011}-2}=\frac{10^{2011}-2+7}{10^{2011}-2}=1+\frac{7}{10^{2011}-2}\)

\(B=\frac{10^{2011}}{10^{2011}-7}=\frac{10^{2011}-7+7}{10^{2011}-7}=1+\frac{7}{10^{2011}-7}\)

Vì \(\frac{7}{10^{2011}-2}< \frac{7}{10^{2011}-7}\Rightarrow1+\frac{7}{10^{2011}-2}< 1+\frac{7}{10^{2011}-7}\Rightarrow A< B\)

m hay lắm Dương, t gửi câu hỏi, m cũng gửi!!! Good Job

(2x-5)-(\(\frac{3}{2}\) . 6x + \(\frac{3}{2}\))=4

2x -5 - 9x -\(\frac{3}{2}\)            =4

2x - 9x                    = 4+ 5+ \(\frac{3}{2}\)

Ta có : 

\(A=1+5+5^2+...+5^{32}\)

\(A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{30}+5^{31}+5^{32}\right)\)

\(A=31+5^3\left(1+5+5^2\right)+...+5^{30}\left(1+5+5^2\right)\)

\(A=31+31.5^3+...+31.5^{30}\)

\(A=31\left(1+5^3+...+5^{30}\right)\) chia hết cho 31 

Vậy \(A\) chia hết cho 31

\(a)\) Ta có : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)

\(\Leftrightarrow\)\(a\left(b+c\right)< b\left(a+c\right)\)

\(\Leftrightarrow\)\(ab+ac< ab+bc\)

\(\Leftrightarrow\)\(ac< bc\)

\(\Leftrightarrow\)\(a< b\)

Mà \(a< b\) \(\Rightarrow\) \(\frac{a}{b}< 1\)

Vậy ...

So sánh : 

Ta có công thức từ câu a) : 

\(\frac{a}{b}< \frac{a+c}{b+c}\) \(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(\frac{10^9+1}{10^{10}+1}< \frac{10^9+1+9}{10^{10}+1+9}=\frac{10^9+10}{10^{10}+10}=\frac{10\left(10^8+1\right)}{10\left(10^9+1\right)}=\frac{10^8+1}{10^9+1}\) ( nhìn phân số đầu với phân số cuối ) 

Vậy \(\frac{10^9+1}{10^{10}+1}< \frac{10^8+1}{10^9+1}\)