so sánh
\(A=\frac{10^{14}-1}{10^{15}-11}\) và \(B=\frac{10^{14}+1}{10^{15}+9}\)
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20 tháng 5 2018
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
7 tháng 2 2023
a: -15/37>-25/37
b: -13/21=-26/42
-9/14=-27/42
mà -26>-42
nên -13/21>-9/14
c: -49/-63=7/9
56/80=7/10
=>-49/-63>56/80
d: 3/14=1-11/14
4/15=1-11/15
mà 11/14>11/15
nên 3/14<4/15
8 tháng 8 2017
Với n > 0 Ta có:
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)
\(=\sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)
\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)
\(\sqrt{16}+\sqrt{9}=3+4=7\)
16 tháng 6 2017
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{10}\)\(+\)\(\frac{1}{10}\)\(-\)\(\frac{1}{11}\)\(+\)\(\frac{1}{11}\)\(-\)\(\frac{1}{12}\)\(+\)\(\frac{1}{12}\)\(-\)\(\frac{1}{13}\)\(+\)\(\frac{1}{13}\)\(-\)\(\frac{1}{14}\)\(+\)\(\frac{1}{14}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{1}{9}\)\(-\)\(\frac{1}{15}\)
\(A\)\(=\)\(\frac{2}{45}\)
16 tháng 6 2017
\(A=\left(\frac{1}{9}.\frac{1}{10}+\frac{1}{10}.\frac{1}{11}\right)+\left(\frac{1}{11}.\frac{1}{12}+\frac{1}{12}.\frac{1}{13}\right)+\left(\frac{1}{13}.\frac{1}{14}+\frac{1}{14}.\frac{1}{15}\right)\)
Sau đó nhân phân phối ra là xong nhé bạn
8 tháng 8 2018
Ta có
\(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)
\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)
\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)
\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)
Lại có \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)
Từ (1),(2) => B>A
12 tháng 6 2018
b, Ta có:\(\dfrac{1+3+3^2+.....+3^{10}}{1+3+3^2+.....+3^9}\) \(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3+3^2+...+3^{10}}{1+3+3^2+...+3^9}\)\(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3.\left(1+3+3^2+...+3^9\right)}{1+3+3^2+...+3^9}\)
\(=\dfrac{1}{1+3+3^2+...+3^9}+3< 4\)
\(\Rightarrow\) \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< 4\) \(\left(1\right)\)
Ta có :\(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)
\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5+5^2+...+5^{10}}{1+5+5^2+....+5^9}\)
\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5.\left(1+5+5^2+...+5^9\right)}{1+5+5^2+...+5^9}\)
\(=\dfrac{1}{1+5+5^2+...+5^9}+5>5\)
\(\Rightarrow\) \(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}>5\) \(\left(2\right)\)
Từ \(\left(1\right)và\left(2\right)\)
\(\Rightarrow\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)
Vậy \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)
12 tháng 6 2018
a, Đặt \(A\)\(=\dfrac{7^{15}}{1+7+7^2+...+7^{14}}\)
\(\Rightarrow\) \(\dfrac{1}{A}\) \(=\dfrac{1+7+7^2+...+7^{14}}{7^{15}}=\dfrac{1}{7^{15}}+\dfrac{7}{7^{15}}+\dfrac{7^2}{7^{15}}+...+\dfrac{7^{14}}{7^{15}}\)
\(=\dfrac{1}{7^{15}}+\dfrac{1}{7^{14}}+\dfrac{1}{7^{13}}+....+\dfrac{1}{7}\)
Đặt \(B=\dfrac{9^{15}}{1+9+9^2+...+9^{14}}\)
\(\Rightarrow\dfrac{1}{B}=\dfrac{1+9+9^2+...+9^{14}}{9^{15}}=\dfrac{1}{9^{15}}+\dfrac{9}{9^{15}}+\dfrac{9^2}{9^{15}}+...+\dfrac{9^{14}}{9^{15}}\)
\(=\dfrac{1}{9^{15}}+\dfrac{1}{9^{14}}+\dfrac{1}{9^{13}}+...+\dfrac{1}{9}\)
Mà \(\dfrac{1}{7^{15}}>\dfrac{1}{9^{15}};\dfrac{1}{7^{14}}>\dfrac{1}{9^{14}};\dfrac{1}{7^{13}}>\dfrac{1}{9^{13}};....;\dfrac{1}{7}>\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{A}>\dfrac{1}{B}\) \(\Rightarrow A< B\)
Vậy\(\dfrac{7^{15}}{1+7+7^2+...+7^{14}}>\dfrac{9^{15}}{1+9+9^2+....+9^{14}}\)
MS
1
Có : 10A = 10^15-10/10^15-11 = (10^15-11)+1/10^15-11 = 1 + 1/10^15-11
10B = 10^15+10/10^15+9 = (10^15+9)+1/10^15+9 = 1 + 1/10^15+9
Vì 10^15-11 < 10^15-9 => 1/10^15-11 > 1/10^15+9 => 10A > 10B
=> A < B
k mk nha