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Bài 1:
a) Ta có: \(13A=\dfrac{13^{16}+13}{13^{16}+1}=1+\dfrac{12}{13^{16}+1}\)
\(13B=\dfrac{13^{17}+13}{13^{17}+1}=1+\dfrac{12}{13^{17}+1}\)
Vì \(\dfrac{12}{13^{16}+1}>\dfrac{12}{13^{17}+1}\Rightarrow1+\dfrac{12}{13^{16}+1}>1+\dfrac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\)
\(\Rightarrow A>B\)
Vậy A > B
b) Ta có: \(1999C=\dfrac{1999^{2000}+1999}{1999^{2000}+1}=1+\dfrac{1998}{1999^{2000}+1}\)
\(1999D=\dfrac{1999^{1999}+1999}{1999^{1999}+1}=1+\dfrac{1998}{1999^{1999}+1}\)
\(\dfrac{1998}{1999^{2000}+1}< \dfrac{1998}{1999^{1999}+1}\Rightarrow1+\dfrac{1998}{1999^{2000}+1}< 1+\dfrac{1999}{1999^{1999}+1}\)
\(\Rightarrow1999C< 1999D\)
\(\Rightarrow C< D\)
Vậy C < D
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
\(\dfrac{-14}{21};\dfrac{-2}{15};\dfrac{14}{-35}\)
\(\dfrac{-17}{21}=\dfrac{-85}{105}\);\(\dfrac{-2}{15}=\dfrac{-14}{105};\dfrac{14}{-35}=\dfrac{-14}{35}=\dfrac{-42}{105}\)
\(\dfrac{17}{60};\dfrac{5}{12};\dfrac{64}{90}\)
\(\dfrac{17}{60}=\dfrac{51}{180};\dfrac{-5}{12}=\dfrac{-75}{180};\dfrac{-64}{90}=\dfrac{-32}{45}=\dfrac{-128}{180}\)
bài2:
a)\(\dfrac{3}{5}>\dfrac{4}{7}\)
b)\(\dfrac{-5}{8}< \dfrac{-7}{12}\)
c)\(\dfrac{5}{-3}< \dfrac{-9}{12}\)
a) \(\dfrac{11}{21}+\dfrac{-4}{7}=\dfrac{11}{21}+\dfrac{-12}{21}=\dfrac{-1}{21}\)
b) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}=\dfrac{1}{3}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}=-1+1+\dfrac{2}{7}=\dfrac{2}{7}\)
c) \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{14}=\dfrac{28}{42}+\dfrac{30}{42}-\dfrac{9}{42}=\dfrac{49}{42}=\dfrac{7}{6}\)
d) \(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}=\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}=\dfrac{14}{35}-\dfrac{15}{35}+\dfrac{7}{35}=\dfrac{6}{35}\)
e) \(\dfrac{21}{47}+\dfrac{9}{45}+\dfrac{26}{47}+\dfrac{45}{5}=\dfrac{21}{47}+\dfrac{1}{5}+\dfrac{26}{47}+\dfrac{45}{5}=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{1}{5}+\dfrac{45}{5}\right)\)
\(=1+\dfrac{46}{5}=\dfrac{51}{5}\)
f) \(\dfrac{15}{12}-\dfrac{18}{13}+\dfrac{5}{13}-\dfrac{3}{12}=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(-\dfrac{18}{13}+\dfrac{5}{13}\right)=1+\left(-1\right)=0\)
g) \(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-4}{9}-\dfrac{5}{9}=\dfrac{-9}{9}=-1\)
h)\(\dfrac{3}{7}+\dfrac{-5}{2}-\dfrac{3}{5}=\dfrac{30}{70}+\dfrac{-175}{70}-\dfrac{42}{70}=\dfrac{-187}{70}\)
i) \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{11.16.3}{12.33.5}=\dfrac{4}{15}\)
a) \(\dfrac{-1}{-4}\)=\(\dfrac{1}{4}>0\)
\(\dfrac{3}{-4}< 0\)
\(\Rightarrow\dfrac{1}{4}>\dfrac{3}{-4}hay\dfrac{-1}{-4}>\dfrac{3}{-4}\)
b) Ta có:
\(\dfrac{15}{17}=1-\dfrac{2}{17}\\ \)
\(\dfrac{25}{27}=1-\dfrac{2}{27}\\ \\ \)
Mà \(\dfrac{2}{17}>\dfrac{2}{27}\left(17< 27\right)\)
\(\Rightarrow1-\dfrac{2}{17}< 1-\dfrac{2}{27}\)hay \(\dfrac{15}{17}< \dfrac{25}{27}\)
Bài 1: Tính ( hợp lý nếu có thể )
\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)
\(=-1+1+\dfrac{2}{-5}\)
\(=0+\dfrac{2}{-5}\)
\(=\dfrac{2}{-5}\)
\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)
\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}\)
\(=\dfrac{2}{3}\)
\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1\)
\(=0\)
\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)
\(=\dfrac{1}{6}+\dfrac{-1}{6}\)
\(=0\)
Bài 2: Tìm x,biết:
a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{2}{3}\)
\(x=\dfrac{2}{15}\)
Vậy \(x=\dfrac{2}{15}\)
b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)
\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=\dfrac{3}{3}=1\)
Vậy \(x=1\)
c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!
\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)
\(x=\dfrac{1}{44}\)
Vậy \(x=\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
a, \(\dfrac{1998}{1999}\) < \(\dfrac{1999}{2000}\)
b, \(\dfrac{47}{15}>\dfrac{29}{35}\)
c, \(\dfrac{12}{25}< \) \(\dfrac{25}{49}\)