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a, (-25) + 276 - (276 - 25)
= (-25) + 276 - 276 + 25
= [ (-25) + 25] - (276 - 276)
= 0 - 0
= 0
b, 24 x 46 + 24 x 53 + 24
= 24 x ( 46 + 53 + 1)
= 24 x 100
= 2400
1. Cho phân số a/b, b>a. C/m :
a) Nếu a/b < 1 thì a+1/b+1 > a/b
b) Nếu a/b > 1 thì a/b > a+1/b+1
2. Tính :
3/4 + 3/12 + 3/24 + 3/40 + 3/60 + 3/84 + 3/112 + 3/144 + 1/180
Toán lớp 6
2A=1/2+1/6+1/12+1/20+1/30+1/42
=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
1: \(\Leftrightarrow x=UCLN\left(24;36;150\right)=6\)
2: \(\Leftrightarrow x\in\left\{24;48;72;...\right\}\)
mà 16<=x<=50
nên \(x\in\left\{24;48\right\}\)
3: \(\Leftrightarrow x\inƯ\left(6\right)\)
mà x>-10
nên \(x\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
4: \(\Leftrightarrow x\in BC\left(4;5;8\right)\)
\(\Leftrightarrow x\in\left\{...;-40;0;40;80;120;160;200;...\right\}\)
mà -20<x<180
nên \(x\in\left\{0;40;80;120;160\right\}\)
Cho tg tren la A
A=\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(A=2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(A=2.\frac{3}{7}\)
\(A=\frac{6}{7}\)
Ta co :
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(=2\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+\frac{1}{8.10}+\frac{1}{10.12}+\frac{1}{12.14}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(=2.\frac{3}{7}\)
\(=\frac{6}{7}\)
Đặt A = 1/4 + ... +1/84
A = 2/8 + 2/24 + ... + 2/168
A = 2/2.4 + 2/4.6 + ... + 2/12.14
A = 1/2 - 1/4 + 1/4 - 1/6 + .. + 1/12 - 1/14
A = 1/2 - 1/14
A = 6/14 = 3/7
\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)
\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)
\(A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)
\(A=\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{12}\right)+\left(\frac{1}{12}-\frac{1}{14}\right)\)
\(A=\frac{1}{2}-\frac{1}{14}\)
\(A=\frac{3}{7}\)
Vậy \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}=\frac{3}{7}\)
a.BCNN(1,8)
1 = 1
8 = 23
BCNN(1,8) = 1.23 = 8
b.BCNN(8,1,12)
8 = 23
1 = 1
12 = 22.3
BCNN(8,1,12)=1.23.3=24
c.BCNN(36,72)
36 = 22.32
72 = 23.32
BCNN(36,72)=23.32= 72
d.BCNN(24,84,180)
24= 23.3
84 = 22.3.7
180=22.32.5
BCNN(24,84,180)= 23.32.5.7=2520
Chúc bạn học tốt ✔
a. BCNN( 1, 8 )
Ta có : 1 = 1
8 = 23
⇒ BCNN( 1, 8 ) = 1 . 23 = 8
b. BCNN( 8, 1, 12 )
Ta có : 8 = 23
1 = 1
12 = 22 . 3
⇒ BCNN( 8, 1, 12 ) = 1 . 23 . 3 = 24
c. BCNN( 36, 72 )
Ta có : 36 = 22 . 32
72 = 23 . 32
⇒ BCNN( 36, 72 ) = 23 . 32 = 72BC
d. BCNN( 24, 84, 180 )
Ta có : 24 = 23 . 3
84 = 22 . 3 . 7
180 = 22 . 32 . 5
⇒ BCNN( 24, 72, 180 ) = 23 . 32 . 5 . 7 = 2520
\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{180}=\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{2}.\left(1-\frac{1}{10}\right)=\frac{1}{2}.\frac{9}{10}=\frac{9}{20}\)