tinh
B= 12+22+32+...+992+1002
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\(x=0-\dfrac{1}{2}\)
\(x=-\dfrac{1}{2}\)
Ta có: \(n+2⋮n-1\)
\(\Leftrightarrow n-1+3⋮n-1\)
\(\Leftrightarrow3⋮n-1\)
Vì \(n\in Z\)\(\Rightarrow n-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow n\in\left\{2;0;4;-2\right\}\)
Vậy....
\(n+2⋮n-1\)
<=> (n-1) +3 \(⋮\) n-1
Vì n-1\(⋮\) n-1
Để n+2\(⋮\) n-1 => 3\(⋮\) n-1 hay n-1\(\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)
=> n \(\in\left\{2;4;0;-2\right\}\)
Bài 1:
a.\(150+50:5-2.3^2\)
\(=150+50:5-2.9\)
\(=150+10-18\)
\(=160-18\)
\(=142\)
b,\(375+693+625+307\)
\(=375+625+\left(693+307\right)\)
\(=1000+1000\)
\(=2000\)
e.\(500-\left\{5\left[409-\left(2^3.3-21\right)^2\right]-1724\right\}\)
\(=500-\left\{5\left[409-\left(8.3-21\right)^2\right]-1724\right\}\)
\(=500-\left\{5\left[409-\left(24-21\right)^2\right]-1724\right\}\)
\(=500-\left\{5\left[409-3^2\right]-1724\right\}\)
\(=500-\left\{5\left[409-9\right]-1724\right\}\)
\(=500-\left\{5.400-1724\right\}\)
\(=500-\left\{2000-1724\right\}\)
\(=500-276\)
\(=224\)
d.\(2^4.5-\left[131-\left(13-4\right)^2\right]\)
\(=16.5-\left[131-11^2\right]\)
\(=80-\left[131-121\right]\)
\(=80-10\)
\(=60\)
c.\(4.2^3-3^4:3^3+25^2:5^2\)
\(=4.8-3^1+625:25\)
\(=32-3+25\)
\(=29+25\)
\(=54\)
Bài 2:
a.\(219-7\left(x+1\right)=100\)
\(7\left(x+1\right)=219-100\)
\(7\left(x+1\right)=119\)
\(x+1=119:7\)
\(x+1=17\)
\(x=17-1\)
\(x=16\)
b.(3x-6)3=36
=>3x-6=36:3
=>3x-6=12
=>3x=12+6
=>3x=18
=>x=18:3
=>x=6
c.716-(x-143)=659
=>x-143=716-659
=>x-143=57
=>x=57+143
=>x=200
d.30-[4(x-2)+15]=3
=>4(x-2)+15=30-3
=>4(x-2)+15=27
=>4(x-2)=27-15
=>4(x-2)=12
=>x-2=12:4
=>x-2=3
=>x=3+2
=>x=5
e.\(\text{[(8x-12):4]3}^3=3^6\)
=>[(8x-12):4]27=729
=>(8x-12):4=729:27
=>(8x-12):4=27
=>8x-12=27.4
=>8x-12=108
=>8x=108+12
=>8x=120
=>x=120:8
=>x=15
gọi số tự nhiên càn tìm là a
Ta có: a:12 dư 5 a-5 chia hết cho 12
a:18 dư 5 =>a-5 chia hết cho 18
a:21 dư 5 a-5 chia hết cho21
=> a-5 thuộc BC(12,18,21)
12=2^2 . 3 =>BCNN(12,18,21)=2^2 .3^2 .7=252
18=2 . 3^2 =>BC(12,18,21)=B(252)=
21=3 . 7
=>BC(12,18,21)=B(252)=B(252)=(...-1008;-756,-504;-252;0;252;504;756,1008,...)
Mà a nhỏ hơn 1000 và lớn hơn 700 => a =756
Vậy a=756
\(B=1^2+2\left(1+1\right)+3\left(2+1\right)+4\left(3+1\right)+100\left(99+1\right)=\)
\(=\left(1+2+3+...+100\right)+\left(1.2+2.3+3.4+...+99.100\right)=\)
1+2+3+...+100 tổng cấp số cộng d=1
\(1.2+2.3+3.4+...+99.100=\)
\(=\dfrac{1}{3}\left(1.2.3+2.3.3+3.4.3+...+99.100.3\right)=\)
\(=\dfrac{1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)}{3}=\)
\(=\dfrac{1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-.....-98.99.100+99.100.101}{3}=\)
\(=\dfrac{99.100.101}{3}=33.100.101\)
Thay các kết quả vào để tính B