So sánh
A=\(\dfrac{10^{15}-9}{10^{15}-3}\)
và B=\(\dfrac{10^{16}-8}{10^{16}-2}\)
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Những câu hỏi liên quan
1 tháng 4 2017
a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)
Vì \(10^{16}+10^{15}>10^{16}+10\)
\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)
Hay A>B
b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)
\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)
Vì \(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)
\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)
Hay C > D
H9
HT.Phong (9A5)
CTVHS
29 tháng 10 2023
a) \(\dfrac{2}{5}-\dfrac{3}{15}\)
\(=\dfrac{2}{5}-\dfrac{3:3}{15:3}\)
\(=\dfrac{2}{5}-\dfrac{1}{5}\)
\(=\dfrac{1}{5}\)
b) \(\dfrac{9}{27}-\dfrac{2}{9}\)
\(=\dfrac{9:3}{27:3}-\dfrac{2}{9}\)
\(=\dfrac{3}{9}-\dfrac{2}{9}\)
\(=\dfrac{1}{9}\)
c) \(\dfrac{18}{24}-\dfrac{4}{8}\)
\(=\dfrac{18:6}{24:6}-\dfrac{4:2}{8:2}\)
\(=\dfrac{3}{4}-\dfrac{2}{4}\)
\(=\dfrac{1}{4}\)
d) \(\dfrac{6}{16}-\dfrac{10}{64}\)
\(=\dfrac{6\times2}{16\times2}-\dfrac{10:2}{64:2}\)
\(=\dfrac{12}{32}-\dfrac{5}{32}\)
\(=\dfrac{7}{32}\)
D
19 tháng 1 2022
2/
a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
19 tháng 1 2022
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
22 tháng 5 2022
a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)
NN
9 tháng 6 2017
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
25 tháng 7 2018
Có:\(10A=\dfrac{10^{16}+10}{10^{16}+1}=\dfrac{10^{16}+1+9}{10^{16}+1}=\dfrac{10^{16}+1}{10^{16}+1}+\dfrac{9}{10^{16}+1}=1+\dfrac{9}{10^{16}+1}\)
\(10B=\dfrac{10^{17}+10}{10^{17}+1}=\dfrac{10^{17}+1+9}{10^{17}+1}=\dfrac{10^{17}+1}{10^{17}+1}+\dfrac{9}{10^{17}+1}=1+\dfrac{9}{10^{17}+1}\)
\(1+\dfrac{9}{10^{16}+1}>1+\dfrac{9}{10^{17}+1}\Rightarrow A>B\)
Vậy \(A>B\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 10 2023
a) \(\dfrac{6}{14}=\dfrac{6:2}{14:2}=\dfrac{3}{7}\)
\(\dfrac{3}{7}< \dfrac{4}{7}\)
b) \(\dfrac{6}{15}=\dfrac{6:3}{15:3}=\dfrac{2}{5}\)
\(\dfrac{3}{5}>\dfrac{2}{5}\)
c) \(\dfrac{10}{18}=\dfrac{10:2}{18:2}=\dfrac{5}{9}\)
\(\dfrac{5}{9}>\dfrac{2}{9}\)
Ta có A = \(\dfrac{10^{15}-3-6}{10^{15}-3}\)= \(\dfrac{10^{15}-3}{10^{15}-3}-\dfrac{6}{10^{15}-3}=1-\dfrac{6}{10^{15}-3}\)
B = \(\dfrac{10^{16}-2-6}{10^{16}-2}=\dfrac{10^{16}-2}{10^{16}-2}-\dfrac{6}{10^{16}-2}\)= \(1-\dfrac{6}{10^{16}-2}\)
Vì \(10^{15}-3\) = \(\overline{100...00}-3=\overline{9...7}\) (1)
\(10^{16}-2=\overline{100...000}-2=\overline{9...8}\) (2)
Từ (1) và (2) =>\(10^{15}-3< 10^{16}-2\) hay \(\dfrac{6}{10^{15}-3}>\dfrac{6}{10^{16}-2}\)
Vậy A > B