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![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(A< 3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đặt \(A=\frac{7^{15}}{1+7+7^2+...+7^{14}}\)
Đặt \(B=1+7+7^2+...+7^{14}\)
\(\Rightarrow7B=7+7^2+...+7^{15}\)
\(\Rightarrow7B-B=6B=7^{15}-1\)
\(\Rightarrow B=\frac{7^{15}-1}{6}\)
\(\Rightarrow A=\frac{7^{15}-1+1}{\frac{7^{15}-1}{6}}=\left(7^{15}-1\right).\frac{6}{7^{15}-1}+\frac{6}{7^{15}-1}=6+\frac{6}{7^{15}-1}\)
Tự làm tiếp nha
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : A = \(\frac{7^{10}}{1+7+7^2+7^3+...+7^9}=1:\frac{1+7+7^2+7^3+...+7^9}{7^{10}}\)
= \(1:\left(\frac{1}{7^{10}}+\frac{7}{7^{10}}+\frac{7^2}{7^{10}}+...+\frac{7^8}{7^{10}}+\frac{7^9}{7^{10}}\right)\)=\(1:\left(\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7^2}+\frac{1}{7}\right)\)
tương tự ta được : B = \(1:\left(\frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5^2}+\frac{1}{5}\right)\)
Vì \(\frac{1}{7^{10}}+\frac{1}{7^9}+\frac{1}{7^8}+...+\frac{1}{7^2}+\frac{1}{7}\)< \(\frac{1}{5^{10}}+\frac{1}{5^9}+\frac{1}{5^8}+...+\frac{1}{5^2}+\frac{1}{5}\)
=> A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) Ta có :
\(\frac{9^{10}-4}{9^{10}-5}=\frac{9^{10}-5+1}{9^{10}-5}=1+\frac{1}{9^{10}-5}\)
\(\frac{9^{10}-2}{9^{10}-3}=\frac{9^{10}-3+1}{9^{10}-3}=1+\frac{1}{9^{10}-3}\)
Do \(\frac{1}{9^{10}-5}>\frac{1}{9^{10}-3}\)
\(\Rightarrow1+\frac{1}{9^{10}-5}>1+\frac{1}{9^{10}-3}\)
\(\Rightarrow\frac{9^{10}-4}{9^{10}-5}>\frac{9^{10}-2}{9^{10}-3}\)
b ) Ta có :
\(\frac{2.7^{10}-1}{7^{10}}=2-\frac{1}{7^{10}}\)
\(\frac{2.7^{10}+1}{7^{10}+1}=\frac{2.7^{10}+2-1}{7^{10}+1}=\frac{2\left(7^{10}+1\right)-1}{7^{10}+1}=2-\frac{1}{7^{10}+1}\)
Do \(\frac{1}{7^{10}}>\frac{1}{7^{10}+1}\)
\(\Rightarrow2-\frac{1}{7^{10}}< 2-\frac{1}{7^{10}+1}\)
\(\Rightarrow\frac{2.7^{10}-1}{7^{10}}< \frac{2.7^{10}+1}{7^{10}+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
xét A và B có: số mũ từ 2 đến 9 giống nhau; mẫu đều cộng 1
=> Ta chỉ có thể so sánh phần cơ số
vì 7>3 => 7 mũ n>3 mũ n
=> A lớn hơn B