Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(100:\left\{250:\left[450-\left(4.5^3-2^5.25\right)\right]\right\}\)
\(=100:\left\{250:\left[450-\left(500-800\right)\right]\right\}\)
\(=100:\left\{250:\left[450+300\right]\right\}\)
\(=100:\left\{250:700\right\}\)
\(=100:\frac{5}{14}\)
\(=280\)
100:{250:[450 - ( 4 . 125 - 4.25 )]}
100: { 250: [ 450 - ( 500 - 100 ) ]}
100: {250:[ 450 - 400]
100: { 250 : 50 }
100: 5 = 20
=( 21999+21999.25): (21990+9)
=21999.(1+25): 21999
=21999.(1+25): 21999
=21999. (1+25): 21999
=1.(1+32)
=1.33
=33
1.
1+2+3+...+99+100
=[(100-1):1+1]x[(100+1):2]
=100x50,5
=5050
2.
a, x2017=x
=> x=1 hoặc x=-1
b, 2x+2=250:8
=> 2x+2=250:23
=> 2x+2=247
=> x+2=47
=> x= 45
c, 3x+3x+2=810
=> 3x+3x+2=34+36
=> x=4
chúc bạn học tốt k mình nha .
\(S=2^0+2^1+2^2+...+2^{99}+2^{100}\)
\(=1+2+\left(2^2+2^3+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=3+2^2.\left(1+2+4\right)+...+2^{98}.\left(1+2+4\right)\)
\(=3+7.\left(2^2+2^5+...+2^{98}\right)\)chia 7 dư 3
\(S=2^0+2^1+2^2+...+2^{99}+2^{100}\)
\(S=\left(2^0+2^1+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{98}+2^{99}+2^{100}\right)\)
\(S=\left(1+2+4\right)+2^3\left(1+2+4\right)+.....+2^{98}\left(1+2+4\right)\)
\(S=7+2^3\cdot7+....+2^{98}\cdot7\)
\(S=7\left(1+2^3+...+2^{98}\right)\)
=> S chia 7 dư 0 hay S chia hết cho 7
a)\(12:\left\{400:\left[500-\left(125+25×7\right)\right]\right\}\)
\(12:\left\{400:\left[500-300\right]\right\}\)
\(12:2\)
\(6\)
b)\(\left[\left(7-3^3:3^2\right):2^2+99\right]-100\)
\(=\left[4:4+99\right]-100\)
\(=100-100\)
\(=0\)
\(c,3^2×\left[\left(5^2-3\right):11\right]-2^4+2×10^3\)
\(=9×2-16+2×10000\)
\(=18-16+20000\)
\(=20002\)
b ) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
= 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
c ) Đặt A = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
=> A < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
=> A < 1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100= 1 - 1/100 = 99/100 < 1
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)< 1
b, \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\)\(\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
c,Ta thấy
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
\(.....\)
\(\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
Sửa đề:
Đặt :
\(A=1^2-2^2+3^2-4^2+5^2-6^2+...+99^2-100^2+101^2\)
\(\Leftrightarrow A=\left(1.1-2.2\right)+\left(3.3-4.4\right)+\left(5.5-6.6\right)+...+\left(99.99-100.100\right)+101.101\)
\(\Leftrightarrow A=\left(-3\right)+\left(-7\right)+\left(-11\right)+...+\left(-199\right)+10201\). Tới đây bạn tìm số số hạng của tổng. Mình tìm được là 50.
\(\Leftrightarrow A=\left[\left(-199\right)+\left(-3\right).50:2\right]+10201=-\left[\left(199+3\right).50:2\right]+10201\)
\(\Leftrightarrow\left(-5050\right)+10201=5151\)
\(2^{100}-\left(1+2+2^2+2^3+...+2^{99}\right)\)
Đặt \(A=\left(1+2+2^2+2^3+...+2^{98}+2^{99}\right)\)
\(2A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)
\(2A-A=2^{100}-1\)
\(A=2^{100}-\left(1+2+2^2+2^3+...+2^{98}+2^{99}\right)\)
\(A=2^{100}-\left(2^{100}-1\right)\)
\(A=1\)
Đặt \(A=1+2+2^2+2^3+...+2^{99};B=2^{100}-\left(1+2+2^2+2^3+...+2^{99}\right)\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{100}\\ \Rightarrow2A-A=\left(2+2^2+2^3+2^4+...+2^{100}\right)-\left(1+2+2^2+2^3+...+2^{99}\right)\\ \Rightarrow A=2^{100}-1\\ \Rightarrow B=2^{100}-A\\ =2^{100}-\left(2^{100}-1\right)\\ =2^{100}-2^{100}+1\\ =1\)