(102+252)-(102-25)2=10n
Tìm n
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Ta có A = \(\dfrac{1919.161616}{323232.3838}=\dfrac{1919.161616}{2.1919.2.161616}=\dfrac{1}{2.2}=\dfrac{1}{4}\)
Vì\(\dfrac{1}{4}=\dfrac{25}{100}\) , mà \(\dfrac{25}{100}>\dfrac{25}{102}=>A>B\)
Ta có : \(A=\dfrac{1919\cdot161616}{323232\cdot3838}=\dfrac{1\cdot1}{2\cdot2}=\dfrac{1}{4}\)
Giả sử \(A=\dfrac{1}{4}=\dfrac{25}{100}\), mà \(\dfrac{25}{100}>\dfrac{25}{102}\)
\(\Rightarrow\) \(\dfrac{1}{4}>\dfrac{25}{102}\)
\(\Rightarrow\) \(A>B.\)
a ) y : 102 = 203 ( dư 101 )
y=203x102+101
y=20 706+101
y=20 807
b ) 25 x y x 4 = 500
25xy=500:4
25xy=125
y=125:25
y=5
a) 172008 = (174)502 = (...1)502 = (....1)
112008 = (....1)
32008 = (34)502 = (...1)502 = (...1)
=> 172008 - 112008 - 32008 = (...1) - (...1) - (...1)
Hiệu 172008 - 112008 tận cùng là 0 => 172008 - 112008 - 32008 tận cùng là 9
b) 1725 = (174)6.17 = (...1)6.17 = (...7)
244 = (242)2 = (...6)2 = (...6)
1321 = (134)5.13 = (...1)5.13 = (...3)
=> B = 1725 - 244 - 1321 = (...7) + (...6) - (....3) = (....0) => B chia hết cho 10
c) Tương tự
8102 - 2102
= 8100.82 - 2100.22
= (84)25.64 - (24)25.4
= (...6)25.64 - (...6)25.4
= (...6).64 - (...6).4
= (...4) - (...4)
= (...0)
8102 - 2102
= 8100.82 - 2100.22
= (84)25.64 - (24)25.4
= (...6)25.64 - (...6)25.4
= (...6).64 - (...6).4
= (...4) - (...4)
= (...0)
a) |5/3 - x| - |-5/6| = |-5/9|
=> |5/3 - x| - 5/6 = 5/9
=> |5/3 - x| = 5/9 + 5/6
=> |5/3 - x| = 25/18
=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)
a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)
\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)