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Bài 1 :
Gọi m là số thứ 151 của dãy ( số cuối cùng )
Ta có : \(m=1+3.\left(151+1\right)=457\)
Ta được :
\(1-4+7-10+13-...+457\)
\(=\)\(\left(1-4\right)+\left(7-10\right)+...+\left(451-454\right)+457\)
\(=\)\(\left(-3\right)+\left(-3\right)+...+\left(-3\right)+457\)
\(=\)\(\left(-3\right).76+457\)
\(=\)\(-228+457\)
\(=\)\(229\)
Bài 2 :
a) Ta có :
\(6\)là bội của \(n+1\)\(\Rightarrow\)\(\left(n+1\right)\inƯ\left(6\right)\)
Mà \(Ư\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
Suy ra : ( ở đây mình lập bảng )
| \(n+1\) | \(1\) | \(-1\) | \(2\) | \(-2\) | \(3\) | \(-3\) | \(6\) | \(-6\) |
| \(n\) | \(0\) | \(-2\) | \(1\) | \(-3\) | \(2\) | \(-4\) | \(5\) | \(-7\) |
Vậy \(n\in\left\{0;-2;1;-3;2;-4;5;-7\right\}\)
b) Ta có :
\(2x-5=2x+2-7=2\left(x+1\right)-7\)chia hết cho \(x+1\)\(\Rightarrow\)\(\left(-7\right)\)chia hết cho \(x+1\)\(\Rightarrow\)\(\left(x+1\right)\inƯ\left(-7\right)\)
Mà \(Ư\left(-7\right)=\left\{1;-1;7;-7\right\}\)
Suy ra : ( mình cũng lập bảng luôn )
| \(x+1\) | \(1\) | \(-1\) | \(7\) | \(-7\) |
| \(x\) | \(0\) | \(-2\) | \(6\) | \(-8\) |
Vậy \(x\in\left\{0;-2;6;-8\right\}\)
Chúc bạn học tốt
Số thứ 151 là :
( 151 + 1 ) . 3 + 1 = 457
Ta có : 1 - 4 + 7 - 10 + . . . + 451 - 454 + 457
=> ( 1 - 4 ) + ( 7 - 10 ) + . . . + ( 451 - 454 ) + 457
=> ( - 3 ) + ( - 3 ) + . . . + ( - 3 ) + 457
=> [ ( - 3 ) . 76 ] + 457
=> ( - 228 ) + 457
=> 229
Ta có:
A = (457/1 + 1) + (456/2 + 1) + ... + (2/456 + 1) + (1/457 + 1) - 457
A = 458 + 458/2 + ... + 458/456 + 458/457 - 457
A = 458 (1 + 1/2 + ...+ 1/456 + 1/457) - 457
Xét 1 + 1/2 + ... + 1/456 + 1/457, ta có
1 = 1
1/2 = 1/2
1/3 + 1/4 > 1/4 + 1/4 = 1/2
1/5 + 1/6 + ... + 1/8 > 1/8 + 1/8 + ... + 1/8 = 1/2
1/9 + 1/10 +...+ 1/16 > 1/16 + 1/16 +...+ 1/16 = 1/2
1/17 + 1/18 + ... + 1/32 > 1/32 + ... + 1/32 = 1/2
1/33+ 1/34 + ... + 1/64 > 1/64 + ...+ 1/64 = 1/2
1/65 + 1/66 + ...+ 1/128 > 1/128 + ... + 1/128 = 1/2
1/129 + 1/130 + ... + 1/256 > 1/256 + ...+ 1/256 = 1/2
1/257 + 1/258 + ... + 1/457 > 1/457 + ... + 1/457 = 201/457 > 0,4
Vậy 1 + 1/2 + ... + 1/456 + 1/457 > 1 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 0,4 = 5,4
Vậy A > 458*5,4 - 457 = 2016,2
Vậy A > 2016.
Ta có:
A = (456/2 + 1) + ... + (2/456 + 1) + (1/457 + 1) + 1
A = 458 + 458/2 + ... + 458/456 + 458/457 - 458/458
A = 458 (1/2 + ...+ 1/456 + 1/457 + 1/458)
Xét 1/2 + ... + 1/456 + 1/458, ta có
1/2 = 1/2
1/3 + 1/4 > 1/4 + 1/4 = 1/2
1/5 + 1/6 + ... + 1/8 > 1/8 + 1/8 + ... + 1/8 = 1/2
1/9 + 1/10 +...+ 1/16 > 1/16 + 1/16 +...+ 1/16 = 1/2
1/17 + 1/18 + ... + 1/32 > 1/32 + ... + 1/32 = 1/2
1/33+ 1/34 + ... + 1/64 > 1/64 + ...+ 1/64 = 1/2
1/65 + 1/66 + ...+ 1/128 > 1/128 + ... + 1/128 = 1/2
1/129 + 1/130 + ... + 1/256 > 1/256 + ...+ 1/256 = 1/2
1/257 + 1/258 + ... + 1/458 > 1/458 + ... + 1/458 = 202/458
Vậy 1/2 + ... + 1/456 + 1/457 > 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + 202/458 = 4 + 202/458 = 2034/458
Vậy A > 458*2034/458 = 2034
Vậy A > 2016.
1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 +......+ 456 - 457
= (1 - 2) + (3 - 4) + (5 - 6) + ...... + (455 - 456) + 457
= -1 + (-1) + (-1) + ...... + (-1) + 457
= -1 x 228 + 457
= -228 + 457
= 229
g: \(=-457+237+23-123=-220-100=-320\)
h: \(=\left(1-3\right)+\left(5-7\right)+...+\left(41-43\right)+\left(45-47\right)\)
\(=\left(-2\right)+\left(-2\right)+...+\left(-2\right)+\left(-2\right)\)
\(=-2\cdot12=-24\)
i: \(=173+27-46-54-19=200-100-19=100-19=81\)
k: \(=-52+82+49-15+13-36\)
\(=30+34-23\)
=30+11
=41
l: \(=\left(3-5\right)+\left(7-9\right)+\left(11-13\right)+\left(15-17\right)\)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)+\left(-2\right)\)
=-8
m: \(=\left(1-2\right)+\left(3-4\right)+...+\left(2001-2002\right)+2003\)
\(=2003-1-1-...-1\)
\(=2003-1001=1002\)
n:Số số hạng là:
\(\left[\left(-51\right)-\left(-99\right)\right]:1+1=49\left(số\right)\)
Tổng là \(\left(-51-99\right)\cdot\dfrac{49}{2}=-3675\)
o: \(=-62-38+1523-2523-92\)
\(=-100+1000-92=900-92=808\)
a) |-45| + |-15| : 3 + |10|.5
= 45 + 15 : 3 + 10.5
= 45 + 5 + 50 = 100
b) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+\frac{3^2}{10.13}+\frac{3^2}{13.16}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\right)\)
\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=3\left(1-\frac{1}{16}\right)=3.\frac{15}{16}=\frac{45}{16}\)
\(A=\left(\dfrac{456}{2}+1\right)+...+\left(\dfrac{2}{456}+1\right)+\left(\dfrac{1}{457}+1\right)+1\)
\(A=458+\dfrac{458}{2}+....+\dfrac{458}{456}+\dfrac{458}{457}-\dfrac{458}{458}\)
\(A=458\left(\dfrac{1}{2}+...+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}\right)\)
Ta xét \(\dfrac{1}{2}+....+\dfrac{1}{456}+\dfrac{1}{457}+\dfrac{1}{458}\)có :
\(\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{3}+\dfrac{1}{4}>\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
\(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{8}>\dfrac{1}{8}+\dfrac{1}{8}+...+\dfrac{1}{8}=\dfrac{1}{2}\)
\(\dfrac{1}{9}+\dfrac{1}{10}+....+\dfrac{1}{16}>\dfrac{1}{16}+....+\dfrac{1}{16}=\dfrac{1}{2}\)
\(\dfrac{1}{17}+\dfrac{1}{18}+....+\dfrac{1}{32}>\dfrac{1}{32}+.....+\dfrac{1}{32}=\dfrac{1}{2}\)
\(\dfrac{1}{33}+\dfrac{1}{34}+....+\dfrac{1}{64}>\dfrac{1}{64}+....+\dfrac{1}{64}=\dfrac{1}{2}\)
\(\dfrac{1}{65}+\dfrac{1}{66}+.....+\dfrac{1}{128}>\dfrac{1}{128}+....+\dfrac{1}{128}=\dfrac{1}{2}\)
\(\dfrac{1}{129}+\dfrac{1}{130}+.....+\dfrac{1}{256}>\dfrac{1}{256}+....+\dfrac{1}{256}=\dfrac{1}{2}\)
\(\dfrac{1}{257}+\dfrac{1}{258}+....+\dfrac{1}{458}>\dfrac{1}{458}+...+\dfrac{1}{458}=\dfrac{1}{2}\)
Vậy ta thấy được rằng
\(\dfrac{1}{2}+...+\dfrac{1}{456}>\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{202}{458}\)
\(=4+\dfrac{202}{458}=\dfrac{2034}{458}\)
Vậy \(A>458.\dfrac{2034}{458}=2034\)
Hay tức là A > 2016 ( đpcm )
Dãy \(1;4;7;10;13;...;451;454;454\)
Có số số hạng là : \(\left(454-1\right):3+1=152\)
Mà gộp thành các cặp nên có số cặp là \(152:2=76\)
Bài này mk có giải rùi mà