viết 20 dm\(^2\) dưới dạng phân số là:
a. \(\dfrac{100}{20}\) m\(^2\)
b. \(\dfrac{20}{100}\) m\(^2\)
c. \(\dfrac{20}{10}\) m\(^2\)
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\(\dfrac{3}{5}m=0,6m\)
\(\dfrac{7}{20}m=0,35m\)
\(\dfrac{5}{4}m=1,25m\)
\(\dfrac{17}{20}m=85cm\)
\(\dfrac{12}{25}m^2=48dm^2\)
\(\dfrac{3}{4}\) ngày = 18 giờ
\(\dfrac{15}{10}=1,5;\dfrac{35}{100}=0,35;\dfrac{107}{100}=1,07\)
\(\dfrac{22109}{1000}=22,109;\dfrac{14}{5}=\dfrac{28}{10}=2,8;\dfrac{920}{1000}=0,92\)
\(\dfrac{138}{100}=1,38;\dfrac{2007}{10}=200,7;\dfrac{1}{1000}=0,001\)
`(1 1/4)^10 . (2/5)^20`
`=(5/4)^10 . (2/5)^20`
`=(5^10 .2^20)/(4^10 .5^20)`
`=(5^10 .4^10)/(4^10 .5^20)`
`=1/(5^10)`
`=(1/5)^10`
Trước hết ta hãy so sánh :
\(\dfrac{10^{100}+1}{10^{101}+1}\)với \(\dfrac{10^{100}+1}{10^{102}+1}\)
Ta có: Cả hai phân số trên cùng tử.
\(\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{102}+1}\)
Tiếp đó so sánh : \(\dfrac{10^{101}+1}{10^{102}+1}\)với \(1\)
Ta được: \(\dfrac{10^{101}+1}{10^{102}+1}< 1\)
Ta lại so sánh được:\(\dfrac{10^{100}+1}{10^{102}+1}< 1\) (*)
Từ (*) suy ra \(\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+2}< \dfrac{10^{101}+1}{10^{102}+1}< 1\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+1}\)
Ngoài ra còn một cách như sau:
\(\dfrac{10^{101}+1}{10^{102}+1}=\dfrac{10^{\left(100+1\right)}+1}{10^{\left(101+1\right)}+1}=\dfrac{10}{10}.\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{101}+1}\) hay B > A hay A < B
Bài 1:
d)
\(\dfrac{x+5}{95}+\dfrac{x+10}{90}+\dfrac{x+15}{85}+\dfrac{x+20}{80}=-4\)
\(\Leftrightarrow\dfrac{x+5}{95}+1+\dfrac{x+10}{90}+1+\dfrac{x+15}{85}+1+\dfrac{x+20}{80}+1=-4+1+1+1+1\)
\(\Leftrightarrow\dfrac{x+100}{95}+\dfrac{x+100}{90}+\dfrac{x+100}{85}+\dfrac{x+100}{80}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\right)=0\)
\(\Leftrightarrow x+100=0\) ( vì: \(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\ne0\))
\(\Leftrightarrow x=-100\)
\(A=\left[\dfrac{1}{100}-1^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{3}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)\(=\left[\dfrac{1}{100}-1^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{3}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)Mà \(\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2=\dfrac{1}{100}-\dfrac{1}{100}=0\)
\(\Rightarrow A=0\)
\(\left(\dfrac{1}{100}-1^2\right)\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)
\(=\left(\dfrac{1}{100}-1^2\right)\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)
\(=\left(\dfrac{1}{100}-1^2\right)\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]...0...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)
\(=0\)
Vậy...
B
B