so sánh 2 số sau: \(\dfrac{-3}{17}\) và \(1\dfrac{7}{10}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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\(\dfrac{25}{100}=\dfrac{10}{40}\)
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
Lời giải:
\(\frac{1}{\sqrt{7}}+\frac{1}{\sqrt{11}}> \frac{1}{\sqrt{4}}+\frac{1}{\sqrt{9}}=\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)
Ta có \(\dfrac{3}{4}=\sqrt{\dfrac{9}{16}}< \sqrt{\dfrac{10}{17}}\Rightarrow\dfrac{3}{4}< \sqrt{\dfrac{10}{17}}\)
\(\sqrt{\dfrac{10}{17}}< \sqrt{\dfrac{9}{16}}\)
mà \(\sqrt{\dfrac{9}{16}}=\dfrac{3}{4}\)
nên \(\sqrt{\dfrac{10}{17}}< \dfrac{3}{4}\)
a: \(\dfrac{-7}{6}=\dfrac{-7\cdot3}{6\cdot3}=\dfrac{-21}{18}\)
\(\dfrac{-11}{9}=\dfrac{-11\cdot2}{9\cdot2}=\dfrac{-22}{18}\)
mà -21>-22
nên \(-\dfrac{7}{6}>-\dfrac{11}{9}\)
b: \(\dfrac{5}{-7}=\dfrac{-5}{7}=\dfrac{-5\cdot5}{7\cdot5}=\dfrac{-25}{35}\)
\(\dfrac{-4}{5}=\dfrac{-4\cdot7}{5\cdot7}=\dfrac{-28}{35}\)
mà -25>-28
nên \(\dfrac{5}{-7}>\dfrac{-4}{5}\)
c: \(\dfrac{-8}{7}< -1\)
\(-1< -\dfrac{2}{5}\)
Do đó: \(-\dfrac{8}{7}< -\dfrac{2}{5}\)
d: \(-\dfrac{2}{5}< 0\)
\(0< \dfrac{1}{3}\)
Do đó: \(-\dfrac{2}{5}< \dfrac{1}{3}\)
a) \(\dfrac{2}{5}=\dfrac{4}{10}\)
\(\dfrac{4}{10}>\dfrac{3}{10}\)
b) \(\dfrac{5}{6}=\dfrac{10}{12}\)
\(\dfrac{7}{12}< \dfrac{10}{12}\)
c) \(\dfrac{1}{2}=\dfrac{2}{4}\)
\(\dfrac{3}{4}< \dfrac{2}{4}\)
d) \(\dfrac{8}{3}=\dfrac{56}{21}\)
\(\dfrac{56}{21}>\dfrac{11}{21}\)
\(\dfrac{-7}{-17}=\dfrac{7}{17}\)
Vì 7>6 nên \(\dfrac{-7}{-17}>\dfrac{6}{17}\)
\(\dfrac{-7}{-17}=\dfrac{7}{17}\)
\(\dfrac{6}{17}\) giữ nguyên
Vì \(7>6\)
\(\Rightarrow\dfrac{-7}{-17}>\dfrac{6}{11}\)
Cách 1:
\(\dfrac{3}{4}=\dfrac{9}{12}\)
\(\dfrac{4}{3}=\dfrac{16}{12}\)
Do đó \(\dfrac{3}{4}< \dfrac{4}{3}\)
Cách 2:
\(\dfrac{3}{4}< 1\)
\(1< \dfrac{4}{3}\)
Do đó \(\dfrac{3}{4}< \dfrac{4}{3}\)
\(-------\)
Cách 1:
\(\dfrac{11}{8}=\dfrac{55}{40}\)
\(\dfrac{7}{10}=\dfrac{28}{40}\)
Do đó \(\dfrac{11}{8}>\dfrac{7}{10}\)
Cách 2:
\(\dfrac{11}{8}>1\)
\(1>\dfrac{7}{10}\)
Do đó \(\dfrac{11}{8}>\dfrac{7}{10}\)
`(-3)/(17)<0`
`1(7)/(10)>0`
`->(-3)/(17)<1(7)/(10)`
Ta có:
\(\dfrac{-3}{17}< 0\)
\(1\dfrac{7}{10}=\dfrac{17}{10}>0\)
Vì \(\dfrac{-3}{17}< 0;\dfrac{17}{10}>0\) nên \(\dfrac{-3}{17}< \dfrac{17}{10}\) hay \(\dfrac{-3}{17}< 1\dfrac{7}{10}\)
Vậy \(\dfrac{-3}{17}< 1\dfrac{7}{10}\)