so sanh \(\frac{29}{33},\frac{22}{37},\frac{29}{37}\)
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\(\frac{22}{37}\) < \(\frac{29}{37}\) < \(\frac{29}{33}\)
Ghi nhớ : Muốn so sánh phân số , ta lấy tử chia cho mẫu
\(\frac{22}{37}\)< \(\frac{29}{37}\)<\(\frac{29}{33}\)
Áp dụng bất đẳng thức : \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\)
Ta chứng minh được \(\frac{20}{39}>\frac{18}{41};\frac{18}{43}>\frac{14}{39};\frac{22}{27}>\frac{22}{29}\)
\(\Rightarrow\frac{20}{39}+\frac{22}{27}+\frac{18}{43}>\frac{14}{37}+\frac{22}{29}+\frac{18}{41}\)
\(\Rightarrow A>B\)
\(\left(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\right)-\left(3\frac{5}{37}-6\frac{14}{29}\right)\)
\(\left(4\frac{5}{37}-3\frac{4}{5}+8\frac{15}{29}\right)-\left(3\frac{5}{37}-6\frac{14}{29}\right)\)
=(1+1+1+1+1+1+1+1)+(1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45)
Đặt A = 1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45
Ta có:
A x 1/2= 1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90
1/6=1/2x3=1/2-1/3
1/12=1/3x4=1/3-1/4
……………………
1/90=1/9x10=1/9-1/10
A x 1/2=1/2-1/3+1/3-1/4+1/4-1/5+…+1/9-1/10
A x 1/2=1/2-1/10=4/10
A=4/10:1/2=4/5
Vậy 4/3+7/6+11/10+16/15+22/21+29/28+37/36+46/45=1+1+1+1+1+1+1+1+4/5=8+4/5=44/5
\(\frac{4}{3}+\frac{7}{6}+\frac{11}{10}+...+\frac{46}{45}\)
\(=1+\frac{1}{3}+1+\frac{1}{6}+1+\frac{1}{10}+...+1+\frac{1}{45}\)
\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)(8 chữ số 1)
\(=8+\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)
Đặt A = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)
=> \(\frac{1}{2}\)A = \(\frac{1}{2}\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)
= \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Vậy A = \(\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)
Do đó biểu thức trên là 8 + \(\frac{4}{5}\) = \(\frac{44}{5}\)
Đáp số: \(\frac{44}{5}\)
1. \(\frac{29}{33}>\frac{29}{37}>\frac{22}{37}\)
2.\(\frac{163}{221}>\frac{163}{257}>\frac{149}{257}\)
22/37<29/37
29/37<29/33
Vậy 22/37<29/37<29/33
Ta có 22/37 < 29/37 và 29/37 < 29/33
=> 22/37< 29/37 < 29/33