* So sánh phân số:
A = \(\dfrac{100^{20}-1}{100^{30}-1}\)
B = \(\dfrac{100^{30}-1}{100^{20}-1}\)
So sánh A=\(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{2021}\)và B=20. So sánh A và B
Có bạn nào giúp mình đc không, nhanh mình sẽ tick cho nhưng nhớ đúng nữa đấy:
Bài 1: So sánh:
A= $\frac{100^{20}-1}{100^{10}-1}$
B= $\frac{100^{30}-1}{100^{20}-1}$
Có bạn nào giúp mình đc không, nhanh mình sẽ tick cho nhưng nhớ đúng nữa đấy:
Bài 1: So sánh:
A= \(\frac{100^{20}-1}{100^{10}-1}\)
B= \(\frac{100^{30}-1}{100^{20}-1}\)
Số?
a) \(\dfrac{36}{42}=\dfrac{18}{?}=\dfrac{?}{7}=\dfrac{30}{?}\) b) \(\dfrac{80}{100}=\dfrac{?}{20}=\dfrac{4}{?}=\dfrac{?}{50}\)
a) \(\dfrac{36}{42}\) = \(\dfrac{18}{21}\) = \(\dfrac{6}{7}\) = \(\dfrac{30}{35}\)
b) \(\dfrac{80}{100}\) =\(\dfrac{16}{20}\) = \(\dfrac{4}{5}\) = \(\dfrac{40}{50}\)
a, 36/42=18/21=6/7=30/35
b, 80/100=16/20=4/5=40/50
x+30/100.x= -1,31
(4,5-2x).(\(-1\dfrac{4}{7}\))=\(\dfrac{11}{4}\)
A=1+2+3+4+5+...+99+100
B=\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+...+\(\dfrac{1}{9900}\)
1) \(x+\dfrac{30}{100}x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}x=-\dfrac{131}{100}\)
\(\Leftrightarrow100x+30x=-131\)
\(\Leftrightarrow130x=-131\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
Vậy \(x=-\dfrac{131}{130}\)
b) \(\left(4,5-2x\right)\cdot\left(-1\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow\left(\dfrac{9}{2}-2x\right)\cdot\left(-\dfrac{4}{7}\right)=\dfrac{11}{4}\)
\(\Leftrightarrow-\dfrac{18}{7}+\dfrac{8}{7}x=\dfrac{11}{4}\)
\(\Leftrightarrow-72+32x=77\)
\(\Leftrightarrow32x=77+72\)
\(\Leftrightarrow32x=149\)
\(\Leftrightarrow x=\dfrac{149}{32}\)
Vậy \(x=\dfrac{149}{32}\)
\(A=1+2+3+4+5+....+99+100\)
\(A=100(100+1):2=5050\)
\(B=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+.....+\dfrac{1}{9900}\)
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}\)
\(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B=1-\dfrac{1}{100}=\dfrac{99}{100}\)
So sánh bt: \(M=\dfrac{100^{100}+1}{100^{99}+1};N=\dfrac{100^{101}+1}{100^{100}+1}\)
Ta có:
\(M=\dfrac{100^{100}+1}{100^{99}+1}\)
\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100\cdot\left(100^{99}+1\right)}\)
\(\Rightarrow\dfrac{M}{100}=\dfrac{100^{100}+1}{100^{100}+100}\)
\(\Rightarrow\dfrac{M}{100}=1-\dfrac{99}{100^{100}+100}\)
\(N=\dfrac{100^{101}+1}{100^{100}+1}\)
\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100\cdot\left(100^{100}+1\right)}\)
\(\Rightarrow\dfrac{N}{100}=\dfrac{100^{101}+1}{100^{101}+100}\)
\(\Rightarrow\dfrac{N}{100}=1-\dfrac{99}{100^{101}+100}\)
Mà: \(100^{101}>100^{100}\)
\(\Rightarrow100^{101}+100>100^{100}+100\)
\(\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}\)
\(\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}\)
\(\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}\)
\(\Rightarrow N< M\)
Tìm các phân số bằng nhau trong các phân số dưới dây:
\(\dfrac{2}{5}\); \(\dfrac{4}{7}\); \(\dfrac{12}{30}\); \(\dfrac{12}{21}\); \(\dfrac{20}{35}\); \(\dfrac{40}{100}\).
2/ So sánh các phân số sau :
a/ \(\dfrac{7}{10}\) và \(\dfrac{11}{15}\) ; b/ \(\dfrac{-1}{8}\) và \(\dfrac{-5}{24}\) ; c/ \(\dfrac{25}{100}\) và \(\dfrac{10}{40}\)
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/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
a)
b)
c) \(\dfrac{25}{100}=\dfrac{10}{40}\)
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\(^2\)