\(\dfrac{100}{100} . . . 1\)
NN
Những câu hỏi liên quan
so sánh:
a)C= \(\dfrac{100^{99}+1}{100^{100}+1}\) và D= \(\dfrac{100^{100}+1}{100^{101}+1}\)
b)E=\(\dfrac{2020^{2021}+1}{2020^{2022}+1}\) và F=\(\dfrac{2020^{2020}+1}{2020^{2021}+1}\)
c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)
\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)
100^100+1<100^101+1
=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)
=>100C>100D
=>C>D
b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)
\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)
2020^2022+1>2020^2021+1(Do 2022>2021)
=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)
=>2020E<2020F
=>E<F
Đúng 0
Bình luận (1)
ND
21 tháng 9 2023 lúc 15:50
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: \(100^{101}>100^{100}\)
\(\Rightarrow100^{101}+100>100^{100}+100\)
\(\Rightarrow\dfrac{99}{100^{101}+100}< \dfrac{99}{100^{100}+100}\)
\(\Rightarrow1-\dfrac{99}{101^{101}+100}< 1-\dfrac{99}{100^{100}+100}\)
\(\Rightarrow\dfrac{N}{100}< \dfrac{M}{100}\)
\(\Rightarrow N< M\)
Đúng 2
Bình luận (0)
giá trị nhỏ nhất của biểu thức left(x+dfrac{1}{3^{ }}right)^2+dfrac{1}{100}làA.dfrac{-1}{2} B.dfrac{1}{100}C.dfrac{-1}{100} D.dfrac{81}{100}GẤP LẮM MN ƠI
Đọc tiếp
giá trị nhỏ nhất của biểu thức \(\left(x+\dfrac{1}{3^{ }}\right)^2+\dfrac{1}{100}\)là
A.\(\dfrac{-1}{2}\) B.\(\dfrac{1}{100}\)
C.\(\dfrac{-1}{100}\) D.\(\dfrac{81}{100}\)
GẤP LẮM MN ƠI
Đúng ghi Đ,sai ghi S:a)dfrac{1}{10} gấp 10 lần dfrac{1}{100} __ b)dfrac{1}{100} gấp 10 lần dfrac{1}{10}__c)dfrac{1}{100} gấp lên 10 lần được dfrac{1}{1000}__ d) dfrac{1}{100} giảm đi 10 lần được dfrac{1}{1000}____ là chỗ điền nha.
Đọc tiếp
Đúng ghi Đ,sai ghi S:
a)\(\dfrac{1}{10}\) gấp 10 lần \(\dfrac{1}{100}\) __ b)\(\dfrac{1}{100}\) gấp 10 lần \(\dfrac{1}{10}\)__
c)\(\dfrac{1}{100}\) gấp lên 10 lần được \(\dfrac{1}{1000}\)__ d) \(\dfrac{1}{100}\) giảm đi 10 lần được \(\dfrac{1}{1000}\)__
__ là chỗ điền nha.
Đúng ghi Đ, sai ghi S
a) dfrac{3}{10} 0,3 .......
dfrac{3}{10}0,3 .......
b)dfrac{135}{100}1,35 ....
dfrac{135}{100}1,35 ........
c) 1dfrac{7}{100}1,7 ......
1dfrac{7}{100} 1,7
Đọc tiếp
Đúng ghi Đ, sai ghi S
a) \(\dfrac{3}{10}< 0,3\) .......
\(\dfrac{3}{10}=0,3\) .......
b)\(\dfrac{135}{100}=1,35\) ....
\(\dfrac{135}{100}>1,35\) ........
c) 1\(\dfrac{7}{100}>1,7\) ......
1\(\dfrac{7}{100}< 1,7\)
( 100 + \(\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}\)) : ( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)) -2
Sửa đề: \(\dfrac{100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)
\(=\dfrac{101\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)
\(=101-2=99\)
Vậy...
Đúng 0
Bình luận (0)
CMR: 100- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\)
Ta có:
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100-1-\dfrac{1}{2}-...-\dfrac{1}{100}=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\)
\(\Rightarrow100=1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+...+\dfrac{1}{100}+\dfrac{99}{100}\)
\(\Rightarrow100=1+1+1+...+1\) (\(100\) số \(1\))
\(\Rightarrow100=100\)
Vậy \(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{99}{100}\) (Đpcm)
Đúng 0
Bình luận (0)
tính D =\(\dfrac{100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....+\dfrac{99}{100}}\)
Ta có :
\(D=\dfrac{100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+.......+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{100-1-\dfrac{1}{2}-\dfrac{1}{3}-......-\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+.....+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{99-\dfrac{1}{2}-\dfrac{1}{3}-......-\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+....+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+.....+\left(1-\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+.......+\dfrac{99}{100}}\)
\(\Leftrightarrow D=\dfrac{\dfrac{1}{2}+\dfrac{2}{3}+........+\dfrac{99}{100}}{\dfrac{1}{2}+\dfrac{2}{3}+......+\dfrac{99}{100}}=1\)
Đúng 0
Bình luận (1)
\(\left(100+\dfrac{99}{2}+\dfrac{98}{3}+\dfrac{97}{4}....+\dfrac{1}{100}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....\dfrac{1}{100}\right)-2\)
CMR 100-(1+\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\))= (\(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}\))
Ta có :
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)\)
\(=100-1-\dfrac{1}{2}-\dfrac{1}{3}-..................-\dfrac{1}{100}\)
\(=99-\dfrac{1}{2}-\dfrac{1}{3}-................-\dfrac{1}{100}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+..................+\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+.................+\dfrac{99}{100}\)
Vậy :\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+....................+\dfrac{99}{100}\)
\(\Rightarrowđpcm\)
Đúng 0
Bình luận (0)