M = (1 + 2 +... + 9)/(11 + 12 +... +19).
Bài 13: Dấu <, =, >
10 … 10 + 3
11 + 2…. 2 + 11
9 … 10 + 9
10 … 10 + 0
17 – 4 … 14 - 3
18 – 4 … 12
15 … 15 – 1
17 + 1… 17 + 2
12+ 5 … 16
16 … 19 - 3
15 – 4 … 10 + 1
19 – 3 … 11
10 < 10 + 3
11 + 2=2 + 11
9 < 10 + 9
10 = 10 + 0
17 – 4 > 14 - 3
18 – 4 >12
15 > 15 – 1
17 + 1<17 + 2
12+ 5 > 16
16 =19 - 3
15 – 4 =10 + 1
19 – 3 >11
M=1/11+1/12=1/13+....=1/19 N=1/3+1/4+1/5+2/7+2/9+2/11
CMR M và N ko phải số nguyên
11+12-1
1+9+10-2-3
11+11+11-19
11+12-1=22
1+9+10-2-3=15
11+11+11-19=14
M = { 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18+ 19 + 20 }
M=210=20 số số hạng
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K MINH NHÉ
M = 210.
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\(M=\left(20+1\right)20:2=210\)
\(\Rightarrow M=210\)
Đưa thừa số chung ra ngoài a, 5/7 . 5/11 + 5/7 . 2/11 - 5/7 . 14/11 b, 19 5/8 : 7/12 - 15 1/4 : 7/12 c, 2/5 . 1/3 - 2/15 : 1/5 + 3/5 . 1/3 d, 4/9 . 19 1/3 - 4/9 . 39 1/3
a: \(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\dfrac{-7}{11}=-\dfrac{5}{11}\)
b: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\left(4+\dfrac{3}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)
c: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2}{15}\cdot\left(-4\right)+\dfrac{3}{15}=\dfrac{-8+3}{15}=\dfrac{-5}{15}=-\dfrac{1}{3}\)
d: \(=\dfrac{4}{9}\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)
1) 1/1×2 + 2/2×4 + 3/4×7 + 4/7×11 +...+ 8/29×37 + 9/37×46 + 10/46×56
2) 4/3×7 + 4/7×11 + 4/11×15 + 4/15×19 + 4/19×23 + 4/23×27
3) 4/3×6 + 4/6×9 + 4/9×12 + 4/12×15 + ... + 4/99×102
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
11+10 19-9
19+2 12-11
20+11 30-11
11+10=21; 19+2=21; 20+11=31: 19-9=10; 12-11=1; 30-11=19
11+10=21; 19+2=21; 20+11=31: 19-9=10; 12-11=1; 30-11=19
sắp xếp các số hữu tỉ sau theo thứ tự tăng dần
a)-12/19;-3/19;-16/19;-1/19;-11/19;-14/19;-9/19
b)-10/9;-10/7;-10/2;-10/4;-10/8;-10/3;-10/11
a, theo thứ tự tăng dần : -16/19;-14/19;-12/19;-11/19;-9/19;-3/19;-1/19
b, theo thứ tự tăng dần :-10/11;-10/9;-10/8;-10/7;-10/4;-10/3;-10/2
a, theo thứ tự tăng dần : -16/19;-14/19;-12/19;-11/19;-9/19;-3/19;-1/19
b, theo thứ tự tăng dần :-10/11;-10/9;-10/8;-10/7;-10/4;-10/3;-10/2
M=9*10+10*11+11*12+12*13+.............+19*20
ta có công thức \(1.2+2.3+3.4+...+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
áp dụng vào bài ta có \(9.10+10.11+11.12+...+19.20=\frac{19.20.21}{3}-\frac{8.9.10}{3}\)
\(=2660-240\)
\(=2420\)