Dien > < =
\(\frac{-3}{4}....\frac{4}{45}\)
kb với mk nha
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\(2\frac{4}{3}+\frac{4}{12}=\frac{2.3+4}{3}+\frac{4}{12}=\frac{10}{3}+\frac{4}{12}=\frac{40}{12}+\frac{4}{12}=\frac{44}{12}=\frac{11}{3}\)
\(2\frac{4}{3}=\frac{10}{3}\)
\(\frac{10}{3}+\frac{4}{12}=\frac{120}{36}+\frac{12}{36}\)
\(=\frac{132}{36}=\frac{22}{6}=\frac{11}{3}\)
\(\frac{3}{12}+\frac{2}{4}-\frac{2}{12}\)
\(=\frac{3}{12}+\frac{6}{12}-\frac{2}{12}\)
\(=\frac{3+6-2}{12}\)
\(=\frac{7}{12}\)
\(\frac{3}{12}+\frac{2}{4}-\frac{2}{12}\)
\(=\frac{3}{12}+\frac{6}{12}-\frac{2}{12}\)
\(=\frac{9}{12}-\frac{2}{12}\)
\(=\frac{7}{12}\)
\(-\frac{3}{4}=-\frac{15}{20}\)
\(\frac{4}{-5}=-\frac{16}{20}\)
Do \(-15>-16\)
\(\Rightarrow-\frac{3}{4}>\frac{4}{-5}\)
So sánh \(\frac{-3}{4}........\frac{4}{-5}\) hay \(\frac{-3}{4}.......\frac{-4}{5}\)
Ta có :
\(1+\frac{-3}{4}=\frac{1}{4}\)
\(1+\frac{-4}{5}=\frac{1}{5}\)
Vì \(\frac{1}{4}>\frac{1}{5}\Rightarrow1+\frac{-3}{4}>1+\frac{-4}{5}\Rightarrow\frac{-3}{4}>\frac{-4}{5}\)
\(\Rightarrow\frac{-3}{4}>\frac{4}{-5}\)
Vậy \(\frac{-3}{4}>\frac{4}{5}\)
= 2x4/3x4 + 3x3/4x3 +3/12=8/12+9/12+3/12=8+9+3/12=20/12=5/3
\(12\frac{4}{5}\)=\(\frac{64}{5}\).
Ta có :
\(\frac{64}{5}+\frac{4}{5}=\frac{68}{5}\)
\(12\frac{4}{5}+\frac{4}{5}=\frac{64}{5}+\frac{4}{5}=\frac{68}{5}\)
1/3+1/4+4/5-1/2=4/12+3/12+4/5-1/2=7/12+4/5-1/2=35/60+48/60-1/2=83/60-1/2=83/60-30/60
=53/60
\(-\frac{3}{4}< \frac{4}{45}\)
Vì : \(-\frac{3}{4}< 1\)
\(\frac{4}{45}>1\)
\(\Rightarrow-\frac{3}{4}< \frac{4}{45}\)
\(\frac{-3}{4}....\frac{4}{45}\)
\(=\frac{-135}{180}....\frac{20}{180}\)
Ta có : - 135 < 20
Nên => \(\frac{-135}{180}< \frac{20}{180}\)
KL : \(\frac{-3}{4}< \frac{4}{45}\)
Cách 2 : So sánh tử với tử , mẫu với mẫu
Ta thấy : - 3 < 4
4 < 45
=> \(\frac{-3}{4}< \frac{4}{45}\)