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\(a)\frac{-3}{4}+\frac{3}{7}+-\frac{1}{4}+\frac{4}{9}+\frac{4}{7}\)
\(=\left(\frac{-3}{4}+-\frac{1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)
\(=-1+1+\frac{4}{9}\)
\(=\frac{4}{9}\)
\(\frac{-3}{4}+\frac{3}{7}+\frac{-1}{4}+\frac{4}{9}+\frac{4}{7}\)
\(=\left(\frac{-3}{4}+\frac{-1}{4}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{4}{9}\)
\(=\left(-1\right)+1+\frac{4}{9}\)
\(=0+\frac{4}{9}\)
\(=\frac{4}{9}\)
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1. \(MSC=BCNN(5,9)=45\)
Quy đồng :
\(\frac{4}{5}=\frac{4\cdot9}{5\cdot9}=\frac{36}{45}\)
\(\frac{6}{9}=\frac{6\cdot5}{9\cdot5}=\frac{30}{45}\)
Mà \(36>30\Rightarrow\frac{36}{45}>\frac{30}{45}\). Vậy : \(\frac{4}{5}>\frac{6}{9}\)
2. \(\frac{9}{7}+\frac{6}{7}=\frac{15}{7}\)
\(\frac{7}{9}+\frac{7}{6}=\frac{42}{54}+\frac{63}{54}=\frac{105}{54}=\frac{35}{18}\)
Tự so sánh :v
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Ta có:
1 = \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+............+\frac{1}{10}\)(10 phân số \(\frac{1}{10}\))
Mà \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};............;\frac{9}{10}>10\)
\(\Rightarrow M>1\)
Vậy M > 1
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\(\frac{1}{7}\cdot\frac{2}{9}+\frac{1}{9}\cdot\frac{3}{7}+\frac{1}{7}\cdot\frac{4}{9}\)
\(=\frac{2}{7}\cdot\frac{1}{9}+\frac{1}{9}\cdot\frac{3}{7}+\frac{4}{7}\cdot\frac{1}{9}\)
\(=\frac{1}{9}\left(\frac{2}{7}+\frac{3}{7}+\frac{4}{7}\right)\)
\(=\frac{1}{9}\cdot\frac{9}{7}=\frac{1}{7}\)
\(\frac{1}{7}.\frac{2}{9}+\frac{1}{9}.\frac{3}{7}+\frac{1}{7}.\frac{4}{9}\)
\(=\frac{2}{7}.\frac{1}{9}+\frac{1}{9}.\frac{3}{7}+\frac{4}{7}.\frac{1}{9}\)
\(=\frac{1}{9}.\left(\frac{2}{7}+\frac{3}{7}+\frac{4}{7}\right)\)
\(=\frac{1}{9}.\frac{9}{7}\)
\(=\frac{1}{7}\)
Nếu tử số = nhau thì ta so sánh mẫu số. Mẫu số nào bé hơn thì lớn hơn
-9/7^4<-9/7^9
Cho mk