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(1+x ) +( 3+x )+(5+x )+(7+x) +....+(97+x)+(99+x)=7050
(x+x+...+x) + ( 1+ 3 +...+ 97+99) = 7050
x.50 + 2500 = 7050
x.50 = 7050 - 2500= 4550
x=4550:50 = 91
Vậy x=91
( 1 + x ) + ( 3 + x ) + ( 5 + x ) + ... + ( 99 + x ) = 7050
(x + x + x + ... + x ) + ( 1 + 3 + 5 + ... + 99 ) = 7050
( x . 50 ) + ( 1 + 99 ) . 50 : 2 = 7050
( x . 50 ) + 2500 = 7050
x . 50 = 7050 - 2500
x . 50 = 4550
x = 4550 : 50
x = 91
Vậy x = 91
học tốt!!!
A= 1 x 3 + 3 x 5 + 5 x7 +...+ 95 x 97 + 97 x 99
\(\Rightarrow6A=1x3x6+3x5x6+5x7x6+...+95x97x6+97x99x6\)
\(6A=1x3x\left(5+1\right)+3x5x\left(7-1\right)+5x7x\left(9-3\right)+...+95x97x\left(99-93\right)+97x99x\left(101-95\right)\)
\(6A=1x3x5+1x3+3x5x7-1x3x5+5x7x9-3x5x7+...+95x97x99-93x95x97\)
\(+97x99x101-95x97x99\)
\(6A=\left(1x3x5+3x5x7+5x7x9+...+95x97x99+97x99x101\right)-\)
\(\left(1x3x5+3x5x7+...+93x95x97+95x97x99\right)+1x3\)
\(6A=97x99x101+1x3\)
\(6A=969903+3\)
\(6A=969906\)
\(A=\frac{969906}{6}\)
\(A=161651\)
1/3*5+1/5*7+1/7*9+...+1/97*99
=1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
=1/3-1/99
32/99
\(53.39+47.39-53.21+47.21\)
\(=39.\left(53-39\right)-\left(53-47\right).21\)
\(=39.14-6.21\)
\(=39.7.2-6.2.7\)
\(=7.2.\left(39-6\right)\)
\(=14.33\)
A=99-97+...+7-5+3-1
=[99-97]+..+[7-5]+[3-1]
=2+...+2+2
=2*50
100
A=99 - 97 + 95 - 93 + 91 - 89 + ... + 7 - 5 + 3 - 1
Ta thấy khoảng cách giữa 2 số liên tiếp là 2
-> Số lượng số hạng của dãy là :(99-1)/2 + 1 =50
Mà cứ 2 số là 1 cặp => có 50/2 =25 cặp tất cả
Vậy A=99 - 97 + 95 - 93 + 91 - 89 + ... + 7 - 5 + 3 - 1
= (99-97)+(95-93)+(91-89)+.....+(7-5)+(3-1)
= 2*25
=50
S = 1 + 3 + 5 + ... + 99
S = (1+99) + (3+97) + ... + ( 49+51)
S= 100 + 100 + ... + 100
S= 25. 100
S = 2500
A = 99 - 97 + 95 - 93 + ... + 7 - 5 + 3 - 1
A = (99-97) + (95-93) + ... + (7-5) + (3-1)
A = 2 + 2 + ... + 2 + 2
A = 2. 25
A= 50
a) \(\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{97.99}\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+\frac{3}{2}.\left(\frac{1}{7}-\frac{1}{9}\right)+...+\frac{3}{2}.\left(\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{32}{99}\)
\(=\frac{16}{33}\)
b)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)
\(=1-\frac{1}{103}\)
\(=\frac{102}{103}\)
\(G=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+.....+\frac{1}{95.97.99}\)
\(=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+....+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(=\frac{1}{4}.\frac{3200}{9603}=\frac{800}{9603}\)
A=1/3x5/3x7/5x9/7x...x97/95x99/97
A=1/3x1/3x99
A=11