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A= 1+7+7^2+7^3+.....+7^100
=) 7A= 7+7^2+7^3+7^4+.....+7^101
=)7A-A=6A=7^101-1
Ta có: 7^101-1 <7^101 =) 6A<B =) A<B
tính
1+5^2+5^4+5^6+...+5^200
GIÚP GIÙM ĐI MÌNH ĐANG CẦN GẤP LẮM
AI NHANH DUNG MINH H CHO
Toán lớp 7
Anh Mai 25/12/2015 lúc 11:46
Đặt A= 1+5^2+5^4+5^6+...+5^200
=> 25A= 5^2+...+5^202
=>25A-A=(5^2+..+5^202)-(1+5^2+..+5^200)
24A=5^202-1
=>
Bài 1:
a) \(\dfrac{-17}{36}\) và \(\dfrac{23}{-48}\)
\(\dfrac{-17}{36}=\dfrac{-17.4}{36.4}=\dfrac{-68}{144}\)
\(\dfrac{23}{-48}=\dfrac{-23}{48}=\dfrac{-23.3}{144.3}=\dfrac{-69}{144}\)
Vì \(\dfrac{-68}{144}>\dfrac{-69}{144}\) nên \(\dfrac{-17}{36}>\dfrac{23}{-48}\)
b) \(\dfrac{-1}{3}\) và \(\dfrac{2}{5}\)
Vì \(\dfrac{-1}{3}\) là số âm mà \(\dfrac{2}{5}\) là số dương nên \(\dfrac{-1}{3}< \dfrac{2}{5}\)
c) \(\dfrac{2}{7}\) và \(\dfrac{5}{4}\)
Vì \(\dfrac{2}{7}< 1\) mà \(\dfrac{5}{4}>1\) nên \(\dfrac{2}{7}< \dfrac{5}{4}\)
d) \(\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\)
\(\dfrac{267}{-268}=\dfrac{-267}{268}=\dfrac{-267.449}{268.449}=\dfrac{-119883}{120332}\)
\(\dfrac{-1347}{1343}=\dfrac{-1347.89}{1343.89}=\dfrac{-119883}{119527}\)
Vì \(\dfrac{-119883}{120332}>\dfrac{-119883}{119527}\) nên \(\dfrac{267}{-268}>\dfrac{-1347}{1343}\)
Bài 2:
\(\dfrac{5}{2}-\left(1\dfrac{3}{7}-0,4\right)=\dfrac{5}{2}-\dfrac{10}{7}-\dfrac{2}{5}=\dfrac{47}{70}\)
Có 207=47.57=47.54.53=47.54.125
có 37.72=37.49
có 47>37
125>49 suy ra 54.125>49
suy ra 47.54.125>37.49
hay 207>37.72
\(\dfrac{-11}{3^7\cdot7^3}=\dfrac{1}{3^7\cdot7^3}\cdot\left(-11\right)\)
\(\dfrac{-78}{3^7\cdot7^4}=\dfrac{-78}{3^7\cdot7^3\cdot7}=\dfrac{1}{3^7\cdot7^3}\cdot\dfrac{-78}{7}\)
mà \(-11>-\dfrac{78}{7}\)
nên \(\dfrac{-11}{3^7\cdot7^3}>\dfrac{-78}{3^7\cdot7^4}\)
-11>-78
nên \(-\dfrac{11}{3^7\cdot7^4}>-\dfrac{78}{3^7\cdot7^4}\)
So sánh 7 và 3^2
3^2 = 3 x 3 = 9
9 > 7
= > 3^2 > 7
có 3^2 = 9
mà 7 < 9
=> 3^2 > 7