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a) Đặt \(A=\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}\)
\(\Rightarrow A=\left(1^2+2^2+..........+100^2\right)\)\(.\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{100.101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+......+100^2\right).\left(1-\frac{1}{101}\right)\)
\(\Rightarrow A=\left(1^2+2^2+.....+100^2\right).\left(\frac{100}{101}\right)\)(a)
Đặt \(M=\left(1^2+2^2+........+100^2\right)\)
\(\Rightarrow M=1.1+2.2+.....+100.100\)
\(\Rightarrow M=1.\left(2-1\right)+2.\left(3-1\right)+....+100.\left(101-1\right)\)
\(\Rightarrow M=\left(1.2-1\right)+\left(2.3-2\right)+.....+\left(100.101-100\right)\)
\(\Rightarrow M=\left(1.2+2.3+.....+100.101\right)-\left(1+2+......+100\right)\)
\(\Rightarrow M=\left(1.2+2.3+......+100.101\right)-5050\)(1)
Đặt \(N=1.2+2.3+....+100.101\)
\(\Rightarrow3.N=1.2.3+2.3.3+......+100.101.3\)
\(\Rightarrow3N=1.2.\left(3-0\right)+2.3.\left(4-1\right)+......+100.101.\left(102-99\right)\)
\(\Rightarrow3N=\left(1.2.3-0\right)+\left(1.2.3-2.3.4\right)+.......+\left(100.101.102-100.101.99\right)\)
\(\Rightarrow3N=100.101.102-0\)
\(\Rightarrow N=343400\)
Thay N = 343400 vào 1) ta được:
M = 343400 - 5050
=> M = 338350
Thay M = 338350 Vào (a) ta được:
A = 338350 . \(\frac{100}{101}\)
=> \(A=\frac{33835000}{101}\)
Vậy \(\frac{1^2}{1.2}+\frac{2^2}{2.3}+.........+\frac{100^2}{100.101}=\frac{33835000}{101}=335000\)
b) Đặt \(B=\frac{2^2}{1.3}+\frac{3^2}{2.4}+..........+\frac{59^2}{58.60}\)
\(\Rightarrow B=\left(2^2+3^2+........+59^2\right).\left(\frac{1}{1.3}+\frac{1}{2.4}+.....+\frac{1}{58.60}\right)\)
Đặt \(G=2^2+3^2+.........+59^2\)VÀ \(H=\frac{1}{1.3}+\frac{1}{2.4}+.........+\frac{1}{58.60}\)
\(\Rightarrow G=2.2+3.3+.......+59.59\) VÀ \(2.H=\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{58.60}\)
Rồi bạn làm như ở phần a) ý
Ta có : A = ( 2 + 22 + 23 ) + ( 24 + 25 + 26 ) + ... + ( 258 + 259 + 260 )
= 2( 1 + 2 + 22 ) + 24 ( 1 + 2 + 22 ) + ... + 258 ( 1 + 2 + 22 )
= 2 ( 1 + 2 + 4 ) + 24 ( 1 + 2 + 4 ) + ... + 258 ( 1 + 2 + 4 )
= 2 x 7 + 24 x 7 + ... + 258 x 7
= 7 x ( 2 + 24 + ... + 258 ) chia hết cho 7
chia hết cho 15 tương tự
a)=407.52.34
=21164.34
=719576
b)=197.(52+23+59)
=197.134
=26398
\(a,\left(4^{19}+3^{21}\right).\left(5^{20}-3^{15}\right).\left(2^6-8^2\right)\\ =\left(4^{19}+3^{21}\right).\left(5^{20}-3^{15}\right).\left(2^{3.2}-2^{3.2}\right)\\ =\left(4^{19}+3^{21}\right).\left(5^{20}-3^{15}\right).0=0\)
\(b,2^5.197+197.3^2+197.59\\ =197.32+197.9+197.59\\ =197.\left(32+9+59\right)\\ =197.100=19700\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Dấu chấm là nhân
a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)
\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)
CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!
a) -4.2.6.25.(-7).5
= (2.5).(-4.25).(-7.6)
= 10.(-100).(-42)
= 42000
b) 16.(38 - 2) - 38.(16 - 1)
= 16.38 - 16.2 - 38.16 - 38
= -38 - 38
= -38.2
= -76.
c) Câu này đề không rõ nha.
A = 2 + 22 + ... + 259
2A = 22 +23 + ... + 260
2A - A = ( 22 +23 + ... + 260 ) - ( 2 + 22 + ... + 259 )
A = 260 - 2
A = 2 + 22 + 23 + 24 +...+ 259
2A = 2 . 2 + 22 . 2 + 23. 2 + 24. 2 +...+ 259 .2
2A = 22 + 23 + 24 + 25 + ... + 260
2A - A = 22 + 23 + 24 + 25 + ... + 260
-
2 + 22 + 23 + 24 +...+ 259
1A = 260 - 2
A = 260 - 2